codeforces Intel Code Challenge Final Round (div.1 + div.2 combined)

比赛水掉3题rk559 rating+115

赛后切掉C

n年没打cf了终于又重新变蓝了，果然太弱。。。

1.A题  Checking the Calendar

　　给定两个星期几，问是否可能分别是两个月的第一天。

　　水题暴力枚举月份

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<string>
#include<cctype>
#include<stack>
#include<queue>
#include<set>
#include<sstream>
#include<map>
using namespace std;
#define FORD(i,k,n) for(int i=n;i>=k;i--)
#define FOR(i,k,n) for(int i=k;i<=n;i++)
#define CLR(a,b) memset(a,b,sizeof(a));
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define LL long long
#define ull unsigned long long
#define i64 long long
#define u32 unsigned int
#define u64 unsigned long long
#define ptb(b,a){int tmp=a;string s;do{s+=tmp%2+‘0‘;tmp/=2;}while(tmp);reverse(s.begin(),s.end());cout<<"bin "<<b<<"="<<s<<endl;}
#define pta(i,a,f,b) {FOR(i,f,b) cout<<a[i]<<" "; printf("\n");}
#define pt(a,b) cout<<a<<"="<<b<<endl
#define pt1(a) cout<<a<<endl
#define pt2(a,b) cout<<a<<" "<<b<<endl
#define pt3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl
#define pt4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl
#define ptl1 cout<<"-------------"<<endl
#define ptl2 cout<<"~~~~~~~~~~~~~~~~"<<endl
#define fp   freopen("in.txt","r",stdin)
#define maxn (100050)
int n,m,k;
int d[4][2]={
    {1,1},{-1,1},{-1,-1},{1,-1}
};
struct node
{
    int x,y;
    ll tm;
    bool operator<(const node&rhs)const
    {
        if(x!=rhs.x)
            return x<rhs.x;
        else if(y!=rhs.y)return y<rhs.y;
        else return tm<rhs.tm;
    }
}p[maxn];
int nume;
node mp[4][maxn*20];
void read()
{
    cin>>n>>m>>k;
    FOR(i,1,k) {
        cin>>p[i].x>>p[i].y;
    }
}
void change(int &x,int &y,int &dir,int& dd)
{
    int nx,ny;
    if(dir==0)
    {
        dd=min(n-x,m-y);
        nx=dd+x,ny=dd+y;
        if(nx==n&&ny==m) ;
        else if(nx==n) dir=1;
        else dir=3;
    }
    else if(dir==1)
    {
        dd=min(x,m-y);
        nx=x-dd,ny=dd+y;
        if(nx==0&&ny==m) ;
        else if(nx==0) dir=0;
        else dir=2;
    }
    else if(dir==2)
    {
        dd=min(x,y);
        nx=x-dd,ny=y-dd;
        if(nx==0&&ny==0) ;
        else if(nx==0) dir=3;
        else dir=1;
    }
    else
    {
        dd=min(n-x,y);
        nx=x+dd,ny=y-dd;
        if(nx==n&&ny==0) ;
        else if(nx==n) dir=2;
        else dir=0;
    }
    x=nx;y=ny;
}
void preprocess()
{
    int x=0,y=0,dir=0;
    ll sum=0;
    int cnt=0;
    while(cnt<=4*(m+n))
    {
        node tmp;tmp.x=x;tmp.y=y;tmp.tm=sum;
        int dd=0;
        mp[dir][nume++]=tmp;
        change(x,y,dir,dd);
        sum+=dd;
        cnt++;
        // ptl2;
    }
    FOR(j,0,3)
        sort(mp[j],mp[j]+nume);
}
void solve()
{
    FOR(i,1,k)
    {
        ll ans=LLINF;
        FOR(j,0,3)
        {
            int x=p[i].x,y=p[i].y,dir=j,dd=0;
            change(x,y,dir,dd);
            node tmp;
            tmp.x=x;tmp.y=y;tmp.tm=0;
            if(j==0) dir=2;
            else if(j==2) dir=0;
            else if(j==1) dir=3;
            else dir=1;
            FOR(t,0,3){
                int p=lower_bound(mp[t],mp[t]+nume,tmp)-mp[t];
                if(p>=0&&p<nume&&mp[t][p].x==x&&mp[t][p].y==y)
                {
                    ans=min(ans,mp[t][p].tm+dd);
                }
            }
        }
        printf("%I64d\n",ans==LLINF?-1:ans);
    }
}
int main()
{

    {
        read();
        preprocess();
        solve();
    }
    return 0;
}

2.B题  Batch Sort

　　给定矩阵要求最后每一行都排好序，支持操作a.每一行最多选两个交换一次 b.最后最多选两列整体交换一次。

　　似乎有很多人在这题上翻船了，估计是想复杂了，因为规模太小暴力枚举b，然后看看a行不行就行了。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<string>
#include<cctype>
#include<stack>
#include<queue>
#include<set>
#include<sstream>
#include<map>
using namespace std;
#define FORD(i,k,n) for(int i=n;i>=k;i--)
#define FOR(i,k,n) for(int i=k;i<=n;i++)
#define CLR(a,b) memset(a,b,sizeof(a));
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define LL long long
#define ull unsigned long long
#define i64 long long
#define u32 unsigned int
#define u64 unsigned long long
#define ptb(b,a){int tmp=a;string s;do{s+=tmp%2+‘0‘;tmp/=2;}while(tmp);reverse(s.begin(),s.end());cout<<"bin "<<b<<"="<<s<<endl;}
#define pta(i,a,f,b) {FOR(i,f,b) cout<<a[i]<<" "; printf("\n");}
#define pt(a,b) cout<<a<<"="<<b<<endl
#define pt1(a) cout<<a<<endl
#define pt2(a,b) cout<<a<<" "<<b<<endl
#define pt3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl
#define pt4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl
#define ptl1 cout<<"-------------"<<endl
#define ptl2 cout<<"~~~~~~~~~~~~~~~~"<<endl
#define fp   freopen("in.txt","r",stdin)
#define maxn 30
int T,N,M,mat[maxn][maxn],tmp[maxn][maxn],a[maxn];
void read()
{
    cin>>N>>M;
    FOR(i,1,N)
    FOR(j,1,M) cin>>mat[i][j];
}
void preprocess()
{

}
int work(int i,int j)
{
    int flg=1;
    FOR(r,1,N)
    {
        swap(tmp[r][i],tmp[r][j]);
    }
    FOR(r,1,N)
    {
        FOR(c,1,M) a[c]=tmp[r][c];
        sort(a+1,a+1+M);
        int cnt=0;
        FOR(c,1,M) if(a[c]!=tmp[r][c]) cnt++;
        if(cnt>2) flg=0;
    }
    return flg;
}
void solve()
{
    int flg=0;
    FOR(i,1,M)
    {
        FOR(j,i,M)
        {
            memcpy(tmp,mat,sizeof(tmp));
            flg|=work(i,j);
        }
    }
    if(flg) puts("YES");
    else puts("NO");
}
int main()
{
    {
        read();
        solve();
    }
    return 0;
}

3.C题 Ray tracing

　　给定一个网格10w*10w，从(0,0)出发光线会反射，询问10w次某一个点能否到达。

　　比赛的时候是想的暴力模拟，记录所有经过的点，mle。于是尝试只记录边界点，tle。

　　正确的思路是暴力只模拟墙上落点，预处理出来。然后每个询问必然光线来自于墙上某个点,于是二分查找有没有这种点，注意用map记录落点会mle。

　　比赛的时候确实有过这种落点思路，但是没想到每个询问必然光线来自于墙上某个点，遗憾。。。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<string>
#include<cctype>
#include<stack>
#include<queue>
#include<set>
#include<sstream>
#include<map>
using namespace std;
#define FORD(i,k,n) for(int i=n;i>=k;i--)
#define FOR(i,k,n) for(int i=k;i<=n;i++)
#define CLR(a,b) memset(a,b,sizeof(a));
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define LL long long
#define ull unsigned long long
#define i64 long long
#define u32 unsigned int
#define u64 unsigned long long
#define ptb(b,a){int tmp=a;string s;do{s+=tmp%2+‘0‘;tmp/=2;}while(tmp);reverse(s.begin(),s.end());cout<<"bin "<<b<<"="<<s<<endl;}
#define pta(i,a,f,b) {FOR(i,f,b) cout<<a[i]<<" "; printf("\n");}
#define pt(a,b) cout<<a<<"="<<b<<endl
#define pt1(a) cout<<a<<endl
#define pt2(a,b) cout<<a<<" "<<b<<endl
#define pt3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl
#define pt4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl
#define ptl1 cout<<"-------------"<<endl
#define ptl2 cout<<"~~~~~~~~~~~~~~~~"<<endl
#define fp   freopen("in.txt","r",stdin)
#define maxn (100050)
int n,m,k;
int d[4][2]={
    {1,1},{-1,1},{-1,-1},{1,-1}
};
struct node
{
    int x,y;
    ll tm;
    bool operator<(const node&rhs)const
    {
        if(x!=rhs.x)
            return x<rhs.x;
        else if(y!=rhs.y)return y<rhs.y;
        else return tm<rhs.tm;
    }
}p[maxn];
int nume;
node mp[4][maxn*20];
void read()
{
    cin>>n>>m>>k;
    FOR(i,1,k) {
        cin>>p[i].x>>p[i].y;
    }
}
void change(int &x,int &y,int &dir,int& dd)
{
    int nx,ny;
    if(dir==0)
    {
        dd=min(n-x,m-y);
        nx=dd+x,ny=dd+y;
        if(nx==n&&ny==m) ;
        else if(nx==n) dir=1;
        else dir=3;
    }
    else if(dir==1)
    {
        dd=min(x,m-y);
        nx=x-dd,ny=dd+y;
        if(nx==0&&ny==m) ;
        else if(nx==0) dir=0;
        else dir=2;
    }
    else if(dir==2)
    {
        dd=min(x,y);
        nx=x-dd,ny=y-dd;
        if(nx==0&&ny==0) ;
        else if(nx==0) dir=3;
        else dir=1;
    }
    else
    {
        dd=min(n-x,y);
        nx=x+dd,ny=y-dd;
        if(nx==n&&ny==0) ;
        else if(nx==n) dir=2;
        else dir=0;
    }
    x=nx;y=ny;
}
void preprocess()
{
    int x=0,y=0,dir=0;
    ll sum=0;
    int cnt=0;
    while(cnt<=4*(m+n))
    {
        node tmp;tmp.x=x;tmp.y=y;tmp.tm=sum;
        int dd=0;
        mp[dir][nume++]=tmp;
        change(x,y,dir,dd);
        sum+=dd;
        cnt++;
        // ptl2;
    }
    FOR(j,0,3)
        sort(mp[j],mp[j]+nume);
}
void solve()
{
    FOR(i,1,k)
    {
        ll ans=LLINF;
        FOR(j,0,3)
        {
            int x=p[i].x,y=p[i].y,dir=j,dd=0;
            change(x,y,dir,dd);
            node tmp;
            tmp.x=x;tmp.y=y;tmp.tm=0;
            if(j==0) dir=2;
            else if(j==2) dir=0;
            else if(j==1) dir=3;
            else dir=1;
            FOR(t,0,3){
                int p=lower_bound(mp[t],mp[t]+nume,tmp)-mp[t];
                if(p>=0&&p<nume&&mp[t][p].x==x&&mp[t][p].y==y)
                {
                    ans=min(ans,mp[t][p].tm+dd);
                }
            }
        }
        printf("%I64d\n",ans==LLINF?-1:ans);
    }
}
int main()
{

    {
        read();
        preprocess();
        solve();
    }
    return 0;
}

4.D题 Dense subsequence

　　给定一个字母序列，要求选定子序列使得原来序列每m个连续子序列中至少包含一个选定子序列的元素。要求字典序最小。

　　从小到大枚举字母。如果尽量少的选当前字母能够达成条件，退出；否则删除这轮选的，全选序列中该字母。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<string>
#include<cctype>
#include<stack>
#include<queue>
#include<set>
#include<sstream>
#include<map>
using namespace std;
#define FORD(i,k,n) for(int i=n;i>=k;i--)
#define FOR(i,k,n) for(int i=k;i<=n;i++)
#define CLR(a,b) memset(a,b,sizeof(a));
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define LL long long
#define ull unsigned long long
#define i64 long long
#define u32 unsigned int
#define u64 unsigned long long
#define ptb(b,a){int tmp=a;string s;do{s+=tmp%2+‘0‘;tmp/=2;}while(tmp);reverse(s.begin(),s.end());cout<<"bin "<<b<<"="<<s<<endl;}
#define pta(i,a,f,b) {FOR(i,f,b) cout<<a[i]<<" "; printf("\n");}
#define pt(a,b) cout<<a<<"="<<b<<endl
#define pt1(a) cout<<a<<endl
#define pt2(a,b) cout<<a<<" "<<b<<endl
#define pt3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl
#define pt4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl
#define ptl1 cout<<"-------------"<<endl
#define ptl2 cout<<"~~~~~~~~~~~~~~~~"<<endl
#define fp   freopen("in.txt","r",stdin)
#define maxn 200000
int T,N,m,flg;
string  ans;
char s[maxn];
int last;
int vis[maxn];
void read()
{
    scanf("%d %s",&m,s+1);
}
void preprocess()
{
}
void solve()
{
    int len=strlen(s+1);
    FOR(i,‘a‘,‘z‘)
    {
        last=0;
        int l=1,r=m;
        FOR(k,1,m) {
            if((last==0&&s[k]==i)||(s[k]<=s[last])||vis[k])
                last=k;
        }
        string tmp;
        if(last&&s[last]==i)
            tmp+=i;
        // pt1(last);
        while(r<len&&last)
        {
            l++;
            r++;
            if(last>=l&&last<=r) continue;
            else
            {
                last=0;
                FOR(k,l,r)
                {
                    if((last==0&&s[k]==i)||(s[k]<=s[last])||vis[k]) last=k;
                }
                if(last&&s[last]==i) {
                    tmp+=i;
                }
            }
            // pt4(last,l,r,ans);
        }
        if(last)
        {
            ans+=tmp;
            break;
        }
        else
        {
            FOR(k,1,len)
            {
                if(s[k]==i) {
                    vis[k]=1;
                    ans+=i;
                }
            }
        }
    }
    cout<<ans<<endl;
}
int main()
{
    {
        read();
        solve();
    }
    return 0;
}

时间： 2024-10-10 16:05:43

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