2021 CCPC Harbin J. local minimum (thinking question)

Problem - J - Codeforces

Ideas ：

A number is the minimum value of its row and column , Then this number is the number to be found . To find out how many such numbers there are in the matrix . You can find the minimum value of each row and column first , If both are a[i][j], be a[i][j] Is a number to be found ,res++

Code ：

#include<iostream>
#include<cstring>
#include<vector>
#include<string>
#include<algorithm>

using namespace std;

int a[1005][1005];
int h[1005], l[1005];
int res;

int main() {

    int n, m;
    scanf("%d%d", &n, &m);

    memset(h, 0x3f, sizeof h);
    memset(l, 0x3f, sizeof l);

    for(int i = 0; i < n; i ++) {
        for(int j = 0; j < m; j ++) {
            scanf("%d", &a[i][j]);
            if(h[i] > a[i][j]) h[i] = a[i][j];
            if(l[j] > a[i][j]) l[j] = a[i][j];
        }
    }

    for(int i = 0; i < n; i ++) {
        for(int j = 0; j < m; j ++) {
            if(a[i][j] == h[i] && a[i][j] == l[j])
            res ++;
        }
    }

    cout << res;
 
    return 0;
}