2018 German Collegiate Programming Contest (GCPC 18)

2018 German Collegiate Programming Contest (GCPC 18)

Attack on Alpha-Zet

建树,求lca

代码:

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl ‘\n‘

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}

/*-----------------------showtime----------------------*/

        const int maxn = 1e3+9;
        string mp[maxn];
        int anc[maxn*maxn][25];
        int fa[maxn*maxn];
        vector<int>tree[maxn*maxn];
        int deep[maxn*maxn];
        int vis[maxn][maxn];

        int n,m;
        void bfs(int sx,int sy){
            queue<pii>que;
            que.push(pii(sx,sy));
            vis[sx][sy] = 1;
            while(!que.empty()){
                int x = que.front().fi,y = que.front().se;
                que.pop();

                int u = (x-1)*m + y;
                if(y<m && mp[x][2*y] != ‘|‘){

                    int tx = x,ty = y+1;
                    if(vis[tx][ty] == 0){
                        vis[tx][ty] = 1;
                        int v = (tx-1)*m+ty;
                        que.push(pii(tx,ty));
                        tree[u].pb(v);
                        tree[v].pb(u);
                    }
                }

                if(y>1&&mp[x][2*y-2] != ‘|‘){
                    int tx = x,ty = y-1;

                    if(vis[tx][ty] == 0){
                        vis[tx][ty] = 1;
                        int v = (tx-1)*m+ty;
                        que.push(pii(tx,ty));
                        tree[u].pb(v);
                        tree[v].pb(u);
                    }
                }

                if(x>1&&mp[x-1][2*y-1] != ‘_‘){
                    int tx = x-1,ty = y;

                    if(vis[tx][ty] == 0){
                        vis[tx][ty] = 1;
                        int v = (tx-1)*m+ty;
                        que.push(pii(tx,ty));
                        tree[u].pb(v);
                        tree[v].pb(u);
                    }

                }

                if(x < n && mp[x][2*y-1] != ‘_‘){
                    int tx = x+1,ty = y;

                    if(vis[tx][ty] == 0){
                        vis[tx][ty] = 1;
                        int v = (tx-1)*m+ty;
                        que.push(pii(tx,ty));
                        tree[u].pb(v);
                        tree[v].pb(u);
                    }

                }
            }
        }

        void dfs(int x){
            anc[x][0] = fa[x];
            for(int i=1; i<=22; i++){
                anc[x][i] = anc[anc[x][i-1]][i-1];
            }
            for(int i=0; i<tree[x].size(); i++){
                if(tree[x][i]!=fa[x]){
                    int y = tree[x][i];
                    fa[y] = x;
                    deep[y] = deep[x]+1;
                    dfs(y);
                }
            }
        }

        int lca(int x,int y){
            int dx = x,dy = y;
            if(deep[x] < deep[y]) swap(x,y);
            for(int i=22; i>=0; i--){
                if(deep[y] <= deep[anc[x][i]])
                    x = anc[x][i];
            }
            int rt;
            if(x==y)rt = x;
            else {
                for(int i=22; i>=0; i--){
                    if(anc[x][i] != anc[y][i]){
                        x = anc[x][i];
                        y = anc[y][i];
                    }
                }
                rt = anc[x][0];
            }
            return deep[dx]+deep[dy] -2* deep[rt];
        }

int main(){
            //scanf("%d%d", &n, &m);
            cin>>n>>m;
            getline(cin,mp[0]);
            for(int i=0; i<=n; i++) getline(cin, mp[i]);

            bfs(1,1);
            dfs(1);

            int q,la;  ll ans = 0;

            scanf("%d", &q);
            for(int i=1; i<=q; i++){
                int x,y;
                scanf("%d%d", &x, &y);
                if(i==1) la = (x-1)*m + y;
                else {
                    int now = (x-1)*m + y;
                    ans += 1ll*lca(la, now);
                    //debug(lca(la,now));
                    la = now;
                }
            }
            printf("%I64d\n", ans);

            return 0;
}

Battle Royale

求两点不经过圆内的最短路径

代码:

#include<bits/stdc++.h>
using namespace std;
#define pi acos(-1.0)

double dis(double x1, double y1, double x2, double y2) {
    return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
}
int main() {
    double x1, x2, y1, y2, ox, oy, r;
    scanf("%lf %lf", &x1, &y1);
    scanf("%lf %lf", &x2, &y2);
    scanf("%lf %lf %lf", &ox, &oy, &r);
    scanf("%lf %lf %lf", &ox, &oy, &r);
    double c = dis(x1, y1, x2, y2);
    double a = dis(ox, oy, x1, y1);
    double b = dis(ox, oy, x2, y2);
    double alpha = acos((a*a + c*c - b*b) / (2*a*c));
    double beta = acos((b*b + c*c - a*a) / (2*b*c));
    if(alpha > pi/2 || beta > pi/2) {
        printf("%.10f\n", c);
        return 0;
    }
    alpha = acos((a*a +b*b - c*c)/ (2*a*b));
    double d = a*b*sin(alpha)/c;
    if(d < r) {
        double aa = acos(r/a);
        double bb = acos(r/b);
        alpha -= aa;
        alpha -= bb;
        printf("%.10f\n", sqrt(a*a - r*r) + sqrt(b*b - r*r) + alpha*r);
    }
    else printf("%.10f\n", c);
    return 0;
}

Coolest Ski Route

按拓扑序转移

代码:

//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl ‘\n‘

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}

/*-----------------------showtime----------------------*/

            const int maxn = 1e3+9;
            vector<pii>mp[maxn];
            queue<int>p,que;
            int n,m;
            int in[maxn];
            int dp[maxn];
            void topo(){
                for(int i=1; i<=n; i++){
                    if(in[i] == 0)p.push(i);
                }
                while(!p.empty()){
                    int u = p.front();p.pop();
                    que.push(u);
                    for(int i=0; i<mp[u].size(); i++){
                        int v = mp[u][i].fi;
                        in[v]--;
                        if(in[v] == 0) p.push(v);
                    }
                }
            }
 int main(){
             scanf("%d%d", &n, &m);

            for(int i=1; i<=m; i++) {
                int u,v,w;
                scanf("%d%d%d", &u, &v, &w);
                mp[u].pb(pii(v,w));
                in[v]++;
            }

            topo();
            int mx = 0;
            while(!que.empty()){
                int u = que.front(); que.pop();
                for(int i=0; i<mp[u].size(); i++){
                    int v = mp[u][i].fi,c = mp[u][i].se;
                    dp[v] = max(dp[v], dp[u] + c);
                    mx = max(mx, dp[v]);
                }
            }
            printf("%d\n", mx);
            return 0;
}

Down the Pyramid

求区间交集

代码:

#include<bits/stdc++.h>
#define pb push_back
#define int long long
using namespace std;
const int maxn=3e5+10;
const int mod=1e9+10+7;
int32_t main()
{
    int n; cin>>n;
    int ans=1e9+10;
    int l=0;
    int r=1e9+10;
    for(int i=1;i<=n;i++)
    {
        int x; cin>>x;
        if(l>x)
        {
            cout<<0<<endl; return 0;
        }
        int q=x-r;
        int p=x-l;
        if(p>x) p=x;
        if(q<0) q=0;
        //cout<<q<<"  "<<p<<endl;
        l=q;
        r=p;
        ans=min(ans,r-l+1);
    }
    cout<<ans<<endl;
}

Expired License

乘1e5再算

代码:

//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl ‘\n‘

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}

/*-----------------------showtime----------------------*/

            const int maxn = 1e5;
            int gcd(int a, int b){
                if(b == 0)return a;
                return gcd(b, a % b);
            }
            bool check(int x){
                if(x == 1) return false;
                for(int i=2; i*i <= x; i++){
                    if(x % i==0)return false;
                }
                return true;
            }

int main(){
         //  cout<<gcd(5 ,10)<<endl;
            int T;  scanf("%d", &T);
            double a,b;
            int n,m;
            while(T--){
                scanf("%lf%lf", &a, &b);
                a += 5e-6;
                b += 5e-6;
                n = (int)(a * maxn );
                m = (int)(b * maxn);

                //cout<<n<<" "<<m<<endl;
                int q = __gcd(n, m);
                n = n/q;
                m = m/q;
               // cout<<n<<" "<<m<<endl;
                if(n == m) printf("2 2\n");
                else if(check(n) && check(m)){
                    printf("%d %d\n", n,m);
                }
                else puts("impossible");
            }

            return 0;
}

Fighting Monsters

代码:

#include<bits/stdc++.h>
#define pb push_back
#define int long long
using namespace std;
const int maxn=1e6+10;
const int mod=1e9+7;
vector<int> vs[maxn];
int32_t main()
{
    int n; cin>>n;
    for(int i=1;i<=n;i++)
    {
        int x; cin>>x; vs[x].push_back(i);
    }
    int l=1;
    int r=1;
    if(vs[1].size()>=2)
    {
        cout<<vs[1][0]<<" "<<vs[1][1]<<endl;
        return 0;
    }
    while(1)
    {
        if(l>1e6||r>1e6)
        {
            break;
        }
        int k=l+r;
        l=r;
        r=k;
        //cout<<l<<" "<<r<<endl;
        if(vs[l].size()&&vs[r].size())
        {
            cout<<vs[l][0]<<" "<<vs[r][0]<<endl;
            return 0;
        }

    }
    cout<<"impossible"<<endl;
}

GPS

Hyper Illuminati

1^(n-1) + 2^(n-1) + ... + s^(n-1) = m

代码:

#include<bits/stdc++.h>
#define pb push_back
#define int long long
using namespace std;
const int maxn=1e6+10;
const int mod=1e9+7;
vector<int> vs[maxn];
int32_t main()
{
    int n; cin>>n;
    for(int i=1;i<=1e6;i++)
    {
        int x=i*(i+1)*(2*i+1)/6;
        if(x<n)
        {
            continue;
        }
        if(n==x)
        {
            cout<<3<<" "<<i<<endl;
            return 0;
        }
        if(x>n)
        {
            break;
        }
    }
    for(int p=4;p<=64;p++)
    {
        int ans=0;
        for(int i=1;i<=1e6;i++)
        {
            int t=1;
            for(int c=1;c<p;c++)
            {
                t*=i;
            }
            ans+=t;
            if(ans<n) continue;
            else if(ans==n)
            {
              cout<<p<<" "<<i<<endl;
              return 0;
            }
            else
            {
                break;
            }
        }
    }
    cout<<"impossible"<<endl;
}

It‘s Time for a Montage

代码:

//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl ‘\n‘

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}

/*-----------------------showtime----------------------*/

        const int maxn = 1e3+9;
        int h[maxn],v[maxn],dif[maxn];
        int n;
        bool check(int x){
            for(int i=1; i<=n; i++){
                if(dif[i] + x > 0) return true;
                else if(dif[i] + x < 0)return false;
            }
            return true;
        }
int main(){
        scanf("%d", &n);
        for(int i=1; i<=n; i++) scanf("%d", &h[i]);
        for(int i=1; i<=n; i++) scanf("%d", &v[i]);
        for(int i=1; i<=n; i++) dif[i] = h[i] - v[i];
        int ans = 0;
        for(int i=0; i<=maxn; i++){
            if(check(i)){ans = i;break;}
        }
        printf("%d\n", ans);

        return 0;

}

Jigsaw Puzzle

模拟

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<long double, long double>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 3e5 + 10;
int a[N][4];
int t[N*2][2];
vector<int> vc;
int main() {
    int n;
    scanf("%d", &n);
    int st = -1;
    int tmp[4];
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < 4; j++) {
            scanf("%d", &a[i][j]);
            if(a[i][j]) {
                int tmp = a[i][j];
                if(t[tmp][0]) t[tmp][1] = i;
                else t[tmp][0] = i;
            }
        }
        swap(a[i][0], a[i][2]);
        for (int j = 0; j < 4; j++) {
            if(!a[i][j] && !a[i][(j+1)%4]) st = i;
        }
    }

    for (int j = 0; j < 4; j++) {
        if(!a[st][j] && !a[st][(j+1)%4]) {
            tmp[0] = tmp[1] = 0;
            tmp[2] = a[st][(j+2)%4];
            tmp[3] = a[st][(j+3)%4];
            break;
        }
    }

    if(~st);
    else return 0*puts("impossible");
    vc.pb(st);
    for (int j = 0; j < 4; j++) a[st][j] = tmp[j];
    while(true) {
        int now = vc.back();
        if(!a[now][2]) break;
        int id = a[now][2];
        int nx;
        if(t[id][0] == now) nx = t[id][1];
        else nx = t[id][0];
        for (int j = 0; j < 4; j++) {
            if(a[nx][j] == id) {
                tmp[0] = a[nx][j];
                tmp[1] = a[nx][(j+1)%4];
                tmp[2] = a[nx][(j+2)%4];
                tmp[3] = a[nx][(j+3)%4];
                break;
            }
        }
        for (int j = 0; j < 4; j++) a[nx][j] = tmp[j];
        if(a[nx][1]) return 0*puts("impossible");
        vc.pb(nx);
    }
    int m = (int)vc.size();
    if(n%m) return 0*puts("impossible");
    if(n == m) {
        for (int i = 0; i < n; i++) if(a[vc[i]][3]) return 0*puts("impossible");
    }
    for (int i = m; i < n; i++) {
        int now = vc[i-m];
        if(!a[now][3]) return 0*puts("impossible");
        int id= a[now][3];
        int nx;
        if(t[id][0] == now) nx = t[id][1];
        else nx = t[id][0];
        for (int j = 0; j < 4; j++) {
            if(a[nx][j] == id) {
                tmp[1] = a[nx][j];
                tmp[2] = a[nx][(j+1)%4];
                tmp[3] = a[nx][(j+2)%4];
                tmp[0] = a[nx][(j+3)%4];
                break;
            }
        }
        for (int j = 0; j < 4; j++) a[nx][j] = tmp[j];
        if(i%m == 0 && a[nx][0]) return 0*puts("impossible");
        if(i%m == m-1 && a[nx][2]) return 0*puts("impossible");
        if(i%m > 0 && (a[nx][0] != a[vc[i-1]][2] || a[nx][0] == 0)) return 0*puts("impossible");
        if(i >= n-m && a[nx][3]) return 0*puts("impossible");
        if(i < n-m && a[nx][3] == 0) return 0*puts("impossible");
        vc.pb(nx);
    }
    printf("%d %d", n/m, m);
    for (int i = 0; i < n; i++) {
        if(i%m == 0) printf("\n");
        printf("%d ", vc[i]);
    }
    return 0;
}

Kitchen Cable Chaos

dp[i][j]:和为i且线段个数为j是否存在

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<long double, long double>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 66, M = 2e3 + 5;
const int INF = 0x3f3f3f3f;
const double eps = 5e-6;
int a[N], n, g;
bool dp[M][N];
int main() {
    scanf("%d %d", &n, &g);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    dp[0][0] = true;
    for (int i = 1; i <= n; i++) {
        for (int j = M-1; j >= a[i]; j--) {
            for (int k = n; k >= 1; k--) {
                dp[j][k] |= dp[j-a[i]][k-1];
            }
        }
    }
    double ans = -1;
    for (int j = g-10; j < M; j++) {
        for (int i = 1; i <= n; i++) {
            if(dp[j][i]) {
                if(j+10-g <= 5*(i+1)) ans = max((j+10.0-g)/(double)(i+1), ans);
            }
        }
    }
    if(ans < -0.5) printf("impossible");
    else printf("%.9f\n", ans);
    return 0;
}

Logic Puzzle

模拟:

代码:

#include<bits/stdc++.h>
using namespace std;
#define pb push_back

const int N = 110;
int a[N][N];
int h, w;
bool vis[N][N];
int dir[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};
int main() {
    scanf("%d %d", &h, &w);
    for (int i = 0; i <= h+1; i++) {
        for (int j = 0; j <= w+1; j++) {
            scanf("%d", &a[i][j]);
        }
    }
    bool f = false;
    for (int i = 0; i <= h+1; i++) {
        for (int j = 0; j <= w+1; j++) {
            int cnt = 0;
            for (int ii = -1; ii <= 1; ii++) {
                for (int jj = -1; jj <= 1; jj++) {
                    int x = i + ii;
                    int y = j + jj;
                    if(1 <= x && x <= h && 1 <= y && y <= w) {
                        if(vis[x][y]) cnt++;
                    }
                }
            }
            int res = a[i][j] - cnt;
            int x = i+1, y = j+1;
            if(1 <= x && x <= h && 1 <= y && y <= w) {
                if(res < 0 || res > 1) {
                    f = true;
                    break;
                }
                else if(res == 1){
                    vis[x][y] = true;
                }
            }
            else {
                if(res != 0) {
                    f = true;
                    break;
                }
            }
        }
    }
    if(f) puts("impossible");
    else {
        for (int i = 1; i <= h; i++) {
            for (int j = 1; j <= w; j++){
                if(vis[i][j]) putchar(‘X‘);
                else putchar(‘.‘);
            }
            puts("");
        }
    }
    return 0;
}

Mountaineers

按高度从小到大启发式合并并查集,合并并查集时更新答案为当前高度

代码:

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define y1 y11
#define pb push_back
#define pii pair<int, int>

const int N = 505, M = 1e6 + 10, MM = 1e5 + 10, NN = N*N;
int fa[NN], sz[NN];
int h[N][N], n, m;
vector<pii> vc[M];
int x1[MM], y1[MM], x2[MM], y2[MM], ans[MM];
set<int> s[NN], que[NN];
map<int, int> mp[NN];
int dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};
void init() {
    for (int i = 0; i < N*N; i++) fa[i] = i, s[i].insert(i);
}
int get_id(int x, int y) {
    return (x-1)*n + y;
}
int Find(int x) {
    if(x == fa[x]) return x;
    else return fa[x] = Find(fa[x]);
}
void Merge(int x, int y, int h) {
    x = Find(x);
    y = Find(y);
    if(x == y) return ;
    if(s[x].size() < s[y].size()) {
        fa[x] = y;
        for(int a : s[x]) {
            s[y].insert(a);
            for (int b : que[a]) {
                if(s[y].find(b) != s[y].end() && mp[a].find(b) == mp[a].end()) {
                    mp[a][b] = h;
                    mp[b][a] = h;
                }
            }
        }
    }
    else {
        fa[y] = x;
        for(int a : s[y]) {
            s[x].insert(a);
            for (int b : que[a]) {
                if(s[x].find(b) != s[x].end() && mp[a].find(b) == mp[a].end()) {
                    mp[a][b] = h;
                    mp[b][a] = h;
                }
            }
        }
    }
}
void solve() {
    for (int i = 1; i < M; i++) {
        for (pii p : vc[i]) {
            int x = p.fi;
            int y = p.se;
            int id = get_id(x, y);
            for (int j = 0; j < 4; j++) {
                int xx = x + dir[j][0];
                int yy = y + dir[j][1];
                if(xx < 1 || xx > m || yy < 1 || yy > n) continue;
                if(h[xx][yy] <= i) {
                    Merge(id, get_id(xx, yy), i);
                }
            }
        }
    }
}
int main() {
    int q;
    scanf("%d %d %d", &m, &n, &q);
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            scanf("%d", &h[i][j]);
            vc[h[i][j]].pb({i, j});
        }
    }
    for (int i = 1; i <= q; i++) {
        scanf("%d %d %d %d", &x1[i], &y1[i], &x2[i], &y2[i]);
        if(x1[i] == x2[i] && y1[i] == y2[i]) {
            ans[i] = h[x1[i]][y1[i]];
        }
        else {
            int id1 = get_id(x1[i], y1[i]);
            int id2 = get_id(x2[i], y2[i]);
            que[id1].insert(id2);
            que[id2].insert(id1);
        }
    }
    init();
    solve();
    for (int i = 1; i <= q; i++) {
        if(ans[i]) printf("%d\n", ans[i]);
        else printf("%d\n", mp[get_id(x1[i], y1[i])][get_id(x2[i], y2[i])]);
    }
    return 0;
}

原文地址:https://www.cnblogs.com/widsom/p/10290561.html

时间: 2024-11-09 03:48:40

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