概率DP ZOJ 3822 Domination

 1 /*
 2     题意：求每行每列都至少有一颗石子的天数期望值
 3     概率DP：先求概率，再算期望，dp[i][j][k]表示一共放了i颗石子，j行k列至少有一颗的概率
 4         由四种状态转移来，行+1， 列+1，行列+1，行列不加。另外，这样写法相当于剪枝，减少时间消耗
 5 */
 6 /************************************************
 7 * Author        :Running_Time
 8 * Created Time  :2015-8-16 19:58:14
 9 * File Name     :D_2.cpp
10  ************************************************/
11 
12 #include <cstdio>
13 #include <algorithm>
14 #include <iostream>
15 #include <sstream>
16 #include <cstring>
17 #include <cmath>
18 #include <string>
19 #include <vector>
20 #include <queue>
21 #include <deque>
22 #include <stack>
23 #include <list>
24 #include <map>
25 #include <set>
26 #include <bitset>
27 #include <cstdlib>
28 #include <ctime>
29 using namespace std;
30 
31 #define lson l, mid, rt << 1
32 #define rson mid + 1, r, rt << 1 | 1
33 typedef long long ll;
34 const int MAXN = 55;
35 const int INF = 0x3f3f3f3f;
36 const int MOD = 1e9 + 7;
37 double dp[MAXN*MAXN][MAXN][MAXN];
38 
39 int main(void)    {     //ZOJ 3822 Domination
40     int T;  scanf ("%d", &T);
41     while (T--) {
42         int n, m;   scanf ("%d%d", &n, &m);
43         memset (dp, 0, sizeof (dp));
44         dp[1][1][1] = 1.0;  int tot = n * m;
45         for (int i=1; i<=tot; ++i)  {
46             for (int j=1; j<=n; ++j)    {
47                 for (int k=1; k<=m; ++k)    {
48                     if (dp[i][j][k] > 0)    {
49                         dp[i+1][j+1][k+1] += dp[i][j][k] / (tot - i) * (n - j) * (m - k);
50                         dp[i+1][j+1][k] += dp[i][j][k] / (tot - i) * (n - j) * k;
51                         dp[i+1][j][k+1] += dp[i][j][k] / (tot - i) * (m - k) * j;
52                         if (j < n || k < m) {
53                             dp[i+1][j][k] += dp[i][j][k] / (tot - i) * (j * k - i);
54                         }
55                     }
56                 }
57             }
58         }
59 
60         double ans = 0.0;
61         for (int i=1; i<=tot; ++i)  ans += dp[i][n][m] * i;
62         printf ("%.10f\n", ans);
63     }
64 
65     return 0;
66 }