Minimax Problem（二分+二进制状态压缩）

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
5
0 3 1 2
8 9 1 3
2 3 4 5
1 0 3 7
3 0 6 3
4 1 7 0
outputCopy
5
思路：最小值最大化问题，绝大概率是二分。但是二分之后的judge环节怎么弄呢？这就是本题的出彩之处。我们将每一个数组中的大于等于mid的位置设置为1，否则就设置为0。这样我们进行或操作，这样出来的就是取最大值操作了。这样如果两两相或，如果各个位上都是1，就代表着当前的mid就是最小值，然后不断的二分，就可以找到最优解了。
代码如下：

#include<bits/stdc++.h>
#define ll long long
using namespace std;

const int maxx=3e5+100;
int a[maxx][10];
int vis[maxx];
int n,m;

inline bool check(int mid,int &x,int &y)
{
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++)
	{
		int cc=0;
		for(int j=1;j<=m;j++) if(a[i][j]>=mid) cc|=(1<<(j-1));
		vis[cc]=i;
	}
	for(int i=0;i<=255;i++)//由于最大就是8位，所以最大是255。
	{
		for(int j=0;j<=255;j++)
		{
			if(vis[i]&&vis[j]&&(i|j)==(1<<m)-1)
			{
				x=vis[i];
				y=vis[j];
				return 1;
			}
		}
	}
	return 0;
}
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=1000000000;
	int x,y;
	while(l<=r)
	{
		int mid=l+r>>1;
		if(check(mid,x,y)) l=mid+1;
		else r=mid-1;
	}
	cout<<x<<" "<<y<<endl;
	return 0;
}

努力加油a啊，(^o)/~