洛谷题目链接

Atcoder题目链接

分析

发现将每个红色连通块涂成绿色连通块后，绿色连通块个数会加一，但是如果这个连通块之前已经跟绿色连通块相邻，则连通块数量减一。

代码

#include <bits/stdc++.h>
#define int long longusing namespace std;const int N = 1005, mod = 998244353;
int n, m, cnt, ans, tot, vis[N][N];
int dx[] = {0, 0, -1, 1}, dy[] = {-1, 1, 0, 0};
char c[N][N];void dfs(int x, int y, int d){if(c[x][y] != '#' || vis[x][y]){return ;}vis[x][y] = d;for(int i = 0; i < 4; i ++){int nx = x + dx[i], ny = y + dy[i];dfs(nx, ny, d);}
}
int qpow(int n, int m, int p){ int res = 1;while(m){if(m & 1){res = res % p * n % p;}n = n % p * n % p;m >>= 1;}return res; 
}signed main(){cin >> n >> m;for(int i = 0; i <= n + 1; i ++){for(int j = 0; j <= m + 1; j ++){c[i][j] = '6';}}for(int i = 1; i <= n; i ++){for(int j = 1; j <= m; j ++){cin >> c[i][j];}}for(int i = 1; i <= n; i ++){for(int j = 1; j <= m; j ++){if(c[i][j] == '#' && !vis[i][j]){cnt ++;dfs(i, j, cnt);}}}for(int i = 1; i <= n; i ++){for(int j = 1; j <= m; j ++){if(c[i][j] == '.'){tot ++;map <int, bool> mp;for(int k = 0; k < 4; k ++){int nx = i + dx[k], ny = j + dy[k];if(c[nx][ny] == '#'){mp[vis[nx][ny]] = true;}}ans += cnt - mp.size() + 1;ans %= mod;}}}cout << ans * qpow(tot, mod - 2, mod) % mod;return 0;
}