1619D - New Year‘s Problem

#include
#include
#include
#include
using namespace std;
const int N = 1e3 + 10;
#define int long long

int n, m;
vector> p;
bool check (int tar) {
	vector abl(m);
	bool pair = 0;
	for (int i = 0; i < n; i ++) {
		int c = 0;
		 for(int j = 0; j < m; ++j){
            if(p[i][j] >= tar){
                abl[j] = true;
                c++;
            }
        }
		if (c > 1) pair = 1;
	}
	if (!pair && m > 1) return 0;
	bool ans = 1;
	for (bool t : abl) ans = ans && t;
	return ans;
}
signed main() {
    int _;
    cin >> _;
    while (_ --) {
    cin >> n >> m;
    p.assign(n, vector(m));
	for (int i = 0;  i< n; i ++) {
		for (int j = 0; j < m; j ++)
		{
		    cin >> p[i][j];
		}
	}	
    	int l = 1, r = 1e9;
    	while(l < r) {
    		int mid = (l + r  + 1)/ 2;
    		if (check(mid)) l = mid;
    		else r = mid - 1;
    	}
    	cout << l << endl;
    }
	return 0;
} 

首先如果能够至少达到x，那么一定至少能够达到x - 1,否则如果不能达到x，那么也不能够达到x + 1