D. Minimax Problem

You are given n arrays a1, a2, ..., an; each array consists of exactly m integers. We denote the y-th element of the x-th array as ax,y. You have to choose two arrays ai and aj (1≤i,j≤n, it is possible that i=j). After that, you will obtain a new array b consisting of m integers, such that for every k∈[1,m] bk=max(ai,k,aj,k). Your goal is to choose i and j so that the value of mink=1mbk is maximum possible. Input The first line contains two integers n and m (1≤n≤3⋅105, 1≤m≤8) — the number of arrays and the number of elements in each array, respectively. Then n lines follow, the x-th line contains the array ax represented by m integers ax,1, ax,2, ..., ax,m (0≤ax,y≤109). Output Print two integers i and j (1≤i,j≤n, it is possible that i=j) — the indices of the two arrays you have to choose so that the value of mink=1mbk is maximum possible. If there are multiple answers, print any of them. Example inputCopy 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 outputCopy 1 5

 

如何获得这个最小值的最大值，暴力必T，可以试着二分。

而二分的check如何判断

check(x)思路，把数组转化成二进制形式，如果当前为大于x,则改位二进制数为1，否则为零，最后的到的二进制数即可替代数组，且他最大为255

然后暴力循环。

 

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<queue>
#define ll long long
const int inf=0x3f3f3f3f;
using namespace std;
const int N=3e5+5;
int vis[260];
int a[N][10];
int n,m;
int x,y;
bool check(int key)
{
    //printf("debug\n");
    int t;
    memset(vis,0,sizeof vis);
    for(int i=1;i<=n;i++)
    {
        t=0;
        for(int j=1;j<=m;j++)
        {
            if(a[i][j]>=key)
            {
                t|=1<<(j-1);
            }
        }
        //printf("%d\n",t);
        vis[t]=i;
    }
    for(int i=0;i<=256;i++)
    {
        for(int j=0;j<=256;j++)
        {
            if(vis[i]&&vis[j]&&((i|j)==((1<<m)-1)))
            {
                x=vis[i];
                y=vis[j];
                //printf("debug\n");
                return true;
            }
        }
    }
    return false;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            scanf("%d",&a[i][j]);
        }
    }
    int l=0,r=1e9;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(check(mid))
        {
            l=mid+1;
        }
        else
            r=mid-1;
    }
    printf("%d %d\n",x,y);
    return 0;
}