CodeForces Gym 102900B Mine Sweeper II

CF 传送门

感觉像脑筋急转弯。

考虑所有数字之和就是相邻的 $(\text{雷}, \text{空地})$ 对数，因此翻转后这个对数不会改变。

然后由于抽屉原理，$b \to a$ 和 $b \to \operatorname{inv}(a)$ 中至少有一个操作次数 $\le \left\lfloor\frac{nm}{2}\right\rfloor$，然后就做完了。

code

// Problem: B. Mine Sweeper II
// Contest: Codeforces - 2020 ICPC Shanghai Site
// URL: https://codeforces.com/gym/102900/problem/B
// Memory Limit: 1024 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)
 
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
 
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;
 
const int maxn = 1010;
 
int n, m;
char a[maxn][maxn], b[maxn][maxn], c[maxn][maxn];
 
void solve() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) {
		scanf("%s", a[i] + 1);
		for (int j = 1; j <= m; ++j) {
			c[i][j] = (a[i][j] == 'X' ? '.' : 'X');
		}
	}
	for (int i = 1; i <= n; ++i) {
		scanf("%s", b[i] + 1);
	}
	int cnt = 0;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			cnt += (a[i][j] != b[i][j]);
		}
	}
	if (cnt <= n * m / 2) {
		for (int i = 1; i <= n; ++i) {
			printf("%s\n", a[i] + 1);
		}
	} else {
		for (int i = 1; i <= n; ++i) {
			printf("%s\n", c[i] + 1);
		}
	}
}
 
int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}