题解

#include <iostream>
#include <cstdio>
#include <memory.h>
using namespace std;
bool mp[105][105], vis[105][105];
int ct, n, m;
bool check(int x, int y){
	int cct = 0;
	for(int i = 1; i <= m; i++){
		if(!vis[i][y]){
			vis[i][y] = 1;
			if(mp[i][y]) cct++;
		}
	}for(int i = 1; i <= n; i++){
		if(!vis[x][i]){
			vis[x][i] = 1;
			if(mp[x][i]) cct++;
		}
	}return cct == ct;
} 
int main(){
	char x;
	cin >> n >> m;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			cin >> x;
			if(x == '*'){
				ct++;
				mp[i][j] = 1;
			}
		}
	}for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			memset(vis, 0, sizeof(vis));
			if(check(i, j)){
				cout << "YES\n" << i << ' ' << j;
				return 0;
			}
		}
	}cout << "NO";

	return 0;
}

时间复杂度：$O(nm(n + m))$