Codeforces

https://codeforc.es/contest/1194/problem/B

好像也没什么思维，就是一个水题，不过蛮有趣的。意思是找缺黑色最少的行列十字。用O(n)的空间预处理掉一维，然后用O(n)的时间根据另一维计算出答案。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
 
int n, m;
string g[50005];
int rq[50005];
 
int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
    //freopen("Yinku.out", "w", stdout);
#endif // Yinku
    int q;
    while(~scanf("%d", &q)) {
        while(q--) {
            scanf("%d%d", &n, &m);
            for(int i = 1; i <= n; i++) {
                cin >> g[i];
                g[i] = '$' + g[i];
            }
            int minans = n * m;
            for(int i = 1; i <= n; i++) {
                rq[i] = 0;
                for(int j = 1; j <= m; j++) {
                    if(g[i][j] == '.') {
                        rq[i]++;
                    }
                }
            }
            for(int j = 1; j <= m; j++) {
                int cq = 0;
                for(int i = 1; i <= n; i++) {
                    if(g[i][j] == '.') {
                        cq++;
                    }
                }
                for(int i = 1; i <= n; i++) {
                    minans = min(minans, rq[i] + cq - (g[i][j] == '.'));
                }
            }
            printf("%d
",minans);
        }
    }
}