(寒假开黑gym)2018 ACM-ICPC, Syrian Collegiate Programming Contest

付队！

许老师！

Hello SCPC 2018! (签到)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const int maxn=1e6+50;
const ll inf=0x3f3f3f3f3f3f3f3fLL;
struct node{
    int value,id;
}a[maxn];
int cmp(node a,node b){
    return a.value<b.value;
}
int f[maxn];
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    freopen("hello.in","r",stdin);
    int t;
    cin>>t;
    while(t--){
        for(int i=1;i<=12;i++)cin>>a[i].value,a[i].id=i;
        sort(a+1,a+1+12,cmp);
        memset(f,0,sizeof(f));
        for(int i=1;i<=4;i++)f[a[i].id]=i;
        int flag=0;
        for(int i=1;i<=4;i++){
            if(f[i]!=i){
                flag=1;
            }
        }
        if(flag)cout<<"no"<<endl;
        else cout<<"yes"<<endl;
    }
    return 0;
}

Binary Hamming (签到)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const int maxn=1e6+50;
const ll inf=0x3f3f3f3f3f3f3f3fLL;
char s[maxn];
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    freopen("hamming.in","r",stdin);
    int t;
    cin>>t;
    while(t--){
        int n,a1=0,b1=0,a2=0,b2=0;
        cin>>n;
        cin>>s;
        for(int i=0;i<n;i++)if(s[i]=='1')a1++;else b1++;
        cin>>s;
        for(int i=0;i<n;i++)if(s[i]=='1')a2++;else b2++;
        cout<<min(a1,b2)+min(b1,a2)<<endl;
    }
    return 0;
}

Portals (模拟,分类讨论)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const int maxn=2e6+50;
const ll inf=0x3f3f3f3f3f3f3f3fLL;
char s[maxn];
int dx[2]={1,-1};
int n;
int S,T;
int dfs(int x,int i){
    if(x==n+1)return 0;
    if(x==0)return 0;
    if(s[x]=='#')return 0;
    if(s[x]=='o')return 1;
    if(s[x]=='s'||s[x]=='e')return 1;
    return dfs(x+dx[i],i);
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    freopen("portals.in","r",stdin);
    int t;
    cin>>t;
    while(t--){
        cin>>n;
        for(int i=0;i<=n+10;i++)s[i]='#';
        cin>>s+1;
        int len=strlen(s+1);
        for(int i=1;i<=n;i++){
            if(s[i]=='s')S=i;
            if(s[i]=='e')T=i;
        }
        int neds=dfs(S+1,0)+dfs(S-1,1);
        int nedt=dfs(T+1,0)+dfs(T-1,1);
        int ans=1e9;
        if(s[S-1]=='o'||s[S+1]=='o'||s[S-1]=='e'||s[S+1]=='e')neds=1e9;
        if(s[T-1]=='o'||s[T+1]=='o'||s[T-1]=='s'||s[T+1]=='s')nedt=1e9;
        ans=min(neds,nedt);
        if(ans==1e9)cout<<-1<<endl;
        else cout<<ans<<endl;
    }
    return 0;
}

Carnival Slots （DP，记忆化搜索）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const int maxn=1e6+50;
const ll inf=0x3f3f3f3f3f3f3f3fLL;
char mp[550][550];

ll dp[505][505];
ll num[550];
ll value[550];
int r,c;

ll dfs(int x,int y){
    if(x==r+1){
        return value[y];
    }
    if(dp[x][y]!=-1)return dp[x][y];
    ll ans=0;
    ans=dfs(x+1,y);

    if(y-1>=1&&mp[x][y]!='.')ans=max(ans,dfs(x+1,y-1));
    if(y+1<=c&&mp[x][y]!='.')ans=max(ans,dfs(x+1,y+1));
    return dp[x][y]=ans;
}

int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    freopen("balls.in","r",stdin);
    int t;
    cin>>t;
    while(t--){
        cin>>r>>c;
        for(int i=1;i<=c;i++)cin>>num[i];
        for(int i=1;i<=r;i++){
            cin>>mp[i]+1;
        }
        for(int i=1;i<=c;i++)cin>>value[i];
        ll ans=0;
        memset(dp,-1,sizeof(dp));
        for(int i=1;i<=c;i++){
            ans+=dfs(1,i)*num[i]*1LL;
        }
        cout<<ans<<endl;
    }
    return 0;
}

Bugged System (模拟)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const int maxn=2e6+50;
const ll inf=0x3f3f3f3f3f3f3f3fLL;
int x[maxn];
struct node{
    int s,d;
}my[maxn];
int in[maxn];
int out[maxn];
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    freopen("bugged.in","r",stdin);
    int t;
    cin>>t;
    while(t--){
        int n,m;
        cin>>n>>m;
        for(int i=1;i<=n;i++)cin>>x[i];
        ll ans=0;
        for(int i=1;i<=m;i++){
            cin>>my[i].s>>my[i].d;
            in[my[i].s]++;
            out[my[i].d]++;
            ans+=abs(x[my[i].s]-x[my[i].d]);
        }
        bool f=0;
        for(int i=1;i<=n;i++){
            if(in[i]!=out[i])f=1;
            in[i]=out[i]=0;
        }
        if(f)cout<<"-1"<<endl;
        else cout<<ans<<endl;
    }
    return 0;
}

Tourists‘ Tour (染色问题，拓扑)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const int maxn=1e6+50;
const ll inf=0x3f3f3f3f3f3f3f3fLL;
vector<int>ve[maxn];
void add(int u,int v){
    ve[v].push_back(u);
    ve[u].push_back(v);
}
int color[maxn];
int vis[10];
int mx;
void dfs(int now){
   // cout<<"now="<<now<<endl;
    memset(vis,0,sizeof(vis));
    for(int i=0;i<ve[now].size();i++){
        vis[color[ve[now][i]]]=1;
    }
    for(int i=1;i<10;i++){
        if(!vis[i]){
            color[now]=i;
            break;
        }
    }
    mx=max(mx,color[now]);
    for(int i=0;i<ve[now].size();i++)if(!color[ve[now][i]])dfs(ve[now][i]);
}
int a[maxn];

int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    freopen("tour.in","r",stdin);
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)cin>>a[i],ve[i].clear();
        stack<int>pq;
        pq.push(1);
        for(int i=2;i<=n;i++){
            while(!pq.empty()&&a[pq.top()]<a[i])pq.pop();
            if(!pq.empty()){
                add(pq.top(),i);
            }
            pq.push(i);
        }
        stack<int>q;
        q.push(n);
        for(int i=n-1;i>=1;i--){
            while(!q.empty()&&a[q.top()]<a[i])q.pop();
            if(!q.empty()){
                add(q.top(),i);
            }
            q.push(i);
        }
        for(int i=1;i<=n;i++)reverse(ve[i].begin(),ve[i].end());
        for(int i=1;i<=n;i++)color[i]=0;
        mx=0;
        for(int i=1;i<=n;i++)if(color[i]==0)dfs(i);
        cout<<mx<<endl;
        for(int i=1;i<=n;i++)cout<<color[i]<<" ";
        cout<<endl;
    }
    return 0;
}

Is Topo Logical? (模拟)

Rise of the Robots (最小圆覆盖)

思路

首先感谢付队的模板！太快了

这题可以把题意转化成求点到所有直线距离的最短距离,然后为什么答案要取反呢，因为我们一开始是以原点为起点的

然后答案给出的是圆心为起点的，所以我们要把这个起点缩回原点。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const ll inf=0x3f3f3f3f3f3f3f3fLL;
#define mp make_pair
const double eps=1e-8;
const double pi=4*atan(1.0);
struct point{
    double x,y;
    point(double a=0,double b=0):x(a),y(b){}
};
int dcmp(double x){ return fabs(x)<eps?0:(x<0?-1:1);}
point operator +(point A,point B) { return point(A.x+B.x,A.y+B.y);}
point operator -(point A,point B) { return point(A.x-B.x,A.y-B.y);}
point operator *(point A,double p){ return point(A.x*p,A.y*p);}
point operator /(point A,double p){ return point(A.x/p,A.y/p);}
point rotate(point A,double rad){
    return point(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
}
bool operator ==(const point& a,const point& b) {
     return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double dot(point A,point B){ return A.x*B.x+A.y*B.y;}
double det(point A,point B){ return A.x*B.y-A.y*B.x;}
double dot(point O,point A,point B){ return dot(A-O,B-O);}
double det(point O,point A,point B){ return det(A-O,B-O);}
double length(point A){ return sqrt(dot(A,A));}
double angle(point A,point B){ return acos(dot(A,B)/length(A)/length(B));}
double distoline(point P,point A,point B)
{
    //点到直线距离
    point v1=B-A,v2=P-A;
    return fabs(det(v1,v2)/length(v1));
}
double distoseg(point P,point A,point B)
{
    //点到线段距离
    if(A==B) return length(P-A);
    point v1=B-A,v2=P-A,v3=P-B;
    if(dcmp(dot(v1,v2))<0) return length(v2);
    else if(dcmp(dot(v1,v3))>0) return length(v3);
    return fabs(det(v1,v2)/length(v1));
}
double Ployarea(vector<point>p)
{
    //多边形面积
    double ans=0; int sz=p.size();
    for(int i=1;i<sz-1;i++) ans+=det(p[i]-p[0],p[i+1]-p[0]);
    return ans/2.0;
}
bool SegmentProperIntersection(point a1,point a2,point b1,point b2)   {
    //规范相交
    double c1=det(a2-a1,b1-a1),c2=det(a2-a1,b2-a1);
    double c3=det(b2-b1,a1-b1),c4=det(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}
bool isPointOnSegment(point p,point a1,point a2)
{
    //点是否在线段上
    return dcmp(det(a1-p,a2-p)==0&&dcmp(dot(a1-p,a2-p))<0);
}
int isPointInPolygon(point p,vector<point>poly)
{
    //判断点与多边形的位置关系
    int wn=0,sz=poly.size();
    for(int i=0;i<sz;i++){
        //在边上
        if(isPointOnSegment(p,poly[i],poly[(i+1)%sz])) return -1;
        int k=dcmp(det(poly[(i+1)%sz]-poly[i],p-poly[i]));
        int d1=dcmp(poly[i].y-p.y); int d2=dcmp(poly[(i+1)%sz].y-p.y);
        if(k>0&&d1<=0&&d2>0) wn++;
        if(k<0&&d2<=0&&d1>0) wn--;
    }
    if(wn!=0) return 1;//内部
    return 0;          //外部
}
double seg(point O,point A,point B){
    if(dcmp(B.x-A.x)==0) return (O.y-A.y)/(B.y-A.y);
    return (O.x-A.x)/(B.x-A.x);
}
pair<double,int>s[110*60];
double polyunion(vector<point>*p,int N){ //有多个才加*,单个不加，有改变的加&
    //求多边形面积并
    double res=0;
    for(int i=0;i<N;i++){
        int sz=p[i].size();
        for(int j=0;j<sz;j++){
            int m=0;
            s[++m]=mp(0,0);
            s[++m]=mp(1,0);
            point a=p[i][j],b=p[i][(j+1)%sz];
            for(int k=0;k<N;k++){
                if(i!=k){
                    int sz2=p[k].size();
                    for(int ii=0;ii<sz2;ii++){
                        point c=p[k][ii],d=p[k][(ii+1)%sz2];
                        int c1=dcmp(det(b-a,c-a));
                        int c2=dcmp(det(b-a,d-a));
                        if(c1==0&&c2==0){
                            if(dcmp(dot(b-a,d-c))){
                                s[++m]=mp(seg(c,a,b),1);
                                s[++m]=mp(seg(c,a,b),-1);
                            }
                        }
                        else{
                            double s1=det(d-c,a-c);
                            double s2=det(d-c,b-c);
                            if(c1>=0&&c2<0) s[++m]=mp(s1/(s1-s2),1);
                            else if(c1<0&&c2>=0) s[++m]=mp(s1/(s1-s2),-1);
                        }
                    }
                }
            }
            sort(s+1,s+m+1);
            double pre=min(max(s[1].first,0.0),1.0),now,sum=0;
            int cov=s[0].second;
            for(int j=2;j<=m;j++){
                now=min(max(s[j].first,0.0),1.0);
                if(!cov) sum+=now-pre;
                cov+=s[j].second;
                pre=now;
            }
            res+=det(a,b)*sum;
        }
    }
    return -(res/2);
}
point jiaopoint(point p,point v,point q,point w)
{   //p+tv q+tw，点加向量表示直线，求直线交点
    point u=p-q;
    double t=det(w,u)/det(v,w);
    return p+v*t;
}
point GetCirPoint(point a,point b,point c)
{
    point p=(a+b)/2;    //ad中点
    point q=(a+c)/2;    //ac中点
    point v=rotate(b-a,pi/2.0),w=rotate(c-a,pi/2.0);   //中垂线的方向向量
    if (dcmp(length(det(v,w)))==0)    //平行
    {
        if(dcmp(length(a-b)+length(b-c)-length(a-c))==0) return (a+c)/2;
        if(dcmp(length(b-a)+length(a-c)-length(b-c))==0) return (b+c)/2;
        if(dcmp(length(a-c)+length(c-b)-length(a-b))==0) return (a+b)/2;
    }
    return jiaopoint(p,v,q,w);
}
point MinCircular(point *P,int n)
{
    //最小圆覆盖 ,看起来是O(N^3)，期望复杂度为O(N)
    random_shuffle(P+1,P+n+1);    //随机化
    point c=P[1]; double r=0;   //c 圆心,,//r 半径
    for (int i=2;i<=n;i++)
        if (dcmp(length(c-P[i])-r)>0)    //不在圆内
        {
            c=P[i],r=0;
            for (int j=1;j<i;j++)
                if (dcmp(length(c-P[j])-r)>0)
                {
                    c=(P[i]+P[j])/2.0;
                    r=length(c-P[i]);
                    for (int k=1;k<j;k++)
                        if (dcmp(length(c-P[k])-r)>0)
                        {
                            c=GetCirPoint(P[i],P[j],P[k]);
                            r=length(c-P[i]);
                        }
                }
        }
    //cout<<r<<":"<<c.x<<" "<<c.y<<endl;
    return c;
}
const int maxn=400010;
bool cmp(point a,point b){ return a.x==b.x?a.y<b.y:a.x<b.x; }
point f[maxn],c[maxn],ch[maxn];
void convexhull(point *a,int n,int &top)
{
    //水平序的Andrew算法求凸包。
    sort(a+1,a+n+1,cmp); top=0;
    for(int i=1;i<=n;i++){ //求下凸包
        while(top>=2&&det(ch[top-1],ch[top],a[i])<=0) top--;
        ch[++top]=a[i];
    }
    int ttop=top;
    for(int i=n-1;i>=1;i--){ //求上凸包
        while(top>ttop&&det(ch[top-1],ch[top],a[i])<=0) top--;
        ch[++top]=a[i];
    }
}
void P(double x)
{
    //if(fabs(x)<0.000001) puts("0.00000000"); else
    printf("%.9lf",x);
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    freopen("robots.in","r",stdin);
    int t;
    cin>>t;
    while(t--){
        int n,R,r;
        cin>>n>>R>>r;
        int tot=1;
        f[tot].x=0;f[tot].y=0;
        for(int i=1;i<=n;i++){
            int a,b;cin>>a>>b;
            tot++;
            f[tot].x=f[tot-1].x+a;
            f[tot].y=f[tot-1].y+b;
        }
        point anw=MinCircular(f,tot);
        cout<<fixed<<setprecision(9);
        cout<<-anw.x<<" "<<-anw.y<<endl;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/luowentao/p/10366989.html