• Educational Codeforces Round 80 D. Minimax Problem


    http://codeforces.com/contest/1288/problem/D

    题意:

    给出一张n*m的表a[n][m],m<=8,从中任选两行i,j

    令b[k]=max(a[i][k],a[j]k])

    最大化 min(b[k])

    二分答案x

    若a[i][j]>=x,令c[i][j]=1,否则c[i][j]=0

    若最终答案>=x,则存在i,j,满足c中第i行和第j行 按位或 的结果都是1

    枚举行肯定超时

    m<=8,每一行的01状态不超过256种,所以可以枚举行的01状态

    可以用bitset做

    #include<cmath>
    #include<cstdio>
    #include<bitset>
    #include<cstring>
    #include<iostream>
    
    using namespace std;
    
    #define N 300001
    
    int n,m,a[N][8];
    bitset<8>b[N];
    
    int vis[256]; 
    int use[256][256],siz[256];
    
    int all;
    
    int turn[N];
    
    bool check(int x)
    {
         memset(vis,0,sizeof(vis));
         for(int i=1;i<=n;++i) 
         {
             b[i].reset();
            for(int j=0;j<m;++j)
                if(a[i][j]>=x) b[i].set(j);
             vis[b[i].to_ulong()]++; 
         }
         for(int i=0;i<=all;++i)
             if(vis[i])
                 for(int j=1;j<=siz[i];++j)
                     if(vis[use[i][j]]) return true;
         return false;
    }
    
    void solve(int x)
    {
         memset(vis,0,sizeof(vis));
         for(int i=1;i<=n;++i) 
         {
             b[i].reset();
            for(int j=0;j<m;++j)
                if(a[i][j]>=x) b[i].set(j);
             vis[b[i].to_ulong()]++; 
             turn[i]=b[i].to_ulong();
         }
         int p=-1,q;
         for(int i=0;i<=all && p==-1;++i)
             if(vis[i])
                 for(int j=1;j<=siz[i];++j)
                     if(vis[use[i][j]])
                     {
                         p=i;
                         q=use[i][j];
                         break;
                    }
         int pp,qq;            
         for(int i=1;i<=n;++i)
         {
             if(turn[i]==p) pp=i;
             if(turn[i]==q) qq=i;
         }
         printf("%d %d",pp,qq);
    } 
    
    int main()
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;++i)
            for(int j=0;j<m;++j)
                scanf("%d",&a[i][j]);
        int l=0,r=1e9,mid,mx;
        all=pow(2,m)-1;
        for(int i=0;i<=all;++i)
             for(int j=i;j<=all;++j)
                 if((i|j)==all) use[i][++siz[i]]=j;
        while(l<=r)
        {
            mid=l+r>>1;
            if(check(mid)) mx=mid,l=mid+1;
            else r=mid-1;
        }
        solve(mx);
        return 0;    
    }
    D. Minimax Problem
    time limit per test
    5 seconds
    memory limit per test
    512 megabytes
    input
    standard input
    output
    standard output

    You are given nn arrays a1a1 , a2a2 , ..., anan ; each array consists of exactly mm integers. We denote the yy -th element of the xx -th array as ax,yax,y .

    You have to choose two arrays aiai and ajaj (1i,jn1≤i,j≤n , it is possible that i=ji=j ). After that, you will obtain a new array bb consisting of mm integers, such that for every k[1,m]k∈[1,m] bk=max(ai,k,aj,k)bk=max(ai,k,aj,k) .

    Your goal is to choose ii and jj so that the value of mink=1mbkmink=1mbk is maximum possible.

    Input

    The first line contains two integers nn and mm (1n31051≤n≤3⋅105 , 1m81≤m≤8 ) — the number of arrays and the number of elements in each array, respectively.

    Then nn lines follow, the xx -th line contains the array axax represented by mm integers ax,1ax,1 , ax,2ax,2 , ..., ax,max,m (0ax,y1090≤ax,y≤109 ).

    Output

    Print two integers ii and jj (1i,jn1≤i,j≤n , it is possible that i=ji=j ) — the indices of the two arrays you have to choose so that the value of mink=1mbkmink=1mbk is maximum possible. If there are multiple answers, print any of them.

    Example
    Input
    Copy
    6 5
    5 0 3 1 2
    1 8 9 1 3
    1 2 3 4 5
    9 1 0 3 7
    2 3 0 6 3
    6 4 1 7 0
    
    Output
    Copy
    1 5
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  • 原文地址:https://www.cnblogs.com/TheRoadToTheGold/p/12219323.html
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