CF1288D Minimax Problem（二分 + 状态压缩）

Minimax Problem

思路：我们通过二分去试探答案，假设答案是x，则我们把数组中大于等于key的标记为1，反之标为0。我们知道m = 8，则01串的情况只有2^8-1种，那么我们可以把n行的情况压缩在vis[2^8-1]长度的数组中，vis记录符合的下标，如果（!vis[x] && !vis[y] && (x | y) == (2^8 -1)）说明这个key是符合的，我们继续通过二分去试探更大的答案。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <functional>
#include <set>
#include <vector>
#include <queue>
#include <cstring>
#include <stack>
 
using namespace std;
 
#define ll long long
#define pb push_back
#define fi first
#define se second
 
vector<vector<int > > a;
vector<int > vis;
int n, m;
int fi, se;

bool check(int mid){
    fill(vis.begin(), vis.end(), 0);
    for(int i = 1; i <= n; ++i){

        int cur = 0;
        for(int j = 1; j <= m; ++j){
            if(a[i][j] >= mid) cur |= (1 << (j - 1));
        }
        //cout << "cur = " << cur << endl;
        vis[cur] = i;
    }

    for(int i = 0; i < (1 << m); ++i){
        for(int j = 0; j < (1 << m); ++j){
            if(vis[i] && vis[j] && (i | j) == (1 << m) - 1){
                fi = vis[i];
                se = vis[j];
                return true;
            }
        }
    }
    return false;
}

void solve(){
    cin >> n >> m;
    a.resize(n + 1, vector<int >(m + 1));
    vis.resize((1 << m) + 10);
    for(int i = 1; i <= n; ++i){
        for(int j = 1; j <= m; ++j)
            cin >> a[i][j];
    }
    int l = 0, r = (int)1e9 + 10;
    while(l <= r){
        int mid = (l + r) >> 1;
        //cout << mid << endl;
        if(check(mid)) l = mid + 1;
        else r = mid - 1;
    }
    cout << fi << " " << se << endl; 
}
 
int main(){
    
    // freopen("C:\Users\admin\Desktop\input.txt", "r", stdin);
    // freopen("C:\Users\admin\Desktop\output.txt", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    solve();
    
    return 0;
}