ZOJ 3822 Domination

来源：互联网发布：淘宝上的手办是真假编辑：程序博客网时间：2024/06/06 23:18

题意：

一个棋盘如果每行每列都有棋子那么这个棋盘达到目标状态现在随机放棋子问达到目标状态的期望步数

思路：

用概率来做计算第k步达到目标状态的概率进而求期望概率计算方法就是dp dp[k][i][j]表示第k步有i行被覆盖j列被覆盖转移只有4种 —— 同时覆盖行列覆盖行覆盖列不覆盖状态数50^4 很简单

代码：

#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<map>#include<set>#include<vector>#include<queue>#include<cstdlib>#include<ctime>#include<cmath>using namespace std;typedef long long LL;#define N 55int t, n, m;double dp[N * N][N][N], ans;int main() {int i, j, k;scanf("%d", &t);while (t--) {memset(dp, 0, sizeof(dp));dp[0][0][0] = 1;ans = 0;scanf("%d%d", &n, &m);for (k = 1; k <= n * m; k++) {for (i = 0; i <= n; i++) {for (j = 0; j <= m; j++) {if (i == n && j == m)break;int f00 = i * j - k + 1;int f01 = i * (m - j);int f10 = (n - i) * j;int f11 = (n - i) * (m - j);int sum = n * m - k + 1;dp[k][i][j] += dp[k - 1][i][j] * f00 / sum;dp[k][i + 1][j] += dp[k - 1][i][j] * f10 / sum;dp[k][i][j + 1] += dp[k - 1][i][j] * f01 / sum;dp[k][i + 1][j + 1] += dp[k - 1][i][j] * f11 / sum;}}ans += dp[k][n][m] * k;}printf("%.10f\n", ans);}return 0;}

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