Zoj 3822 probability dp

Zoj 3822 probability dp

Domination

Time Limit: 8 Seconds Memory Limit: 131072 KB Special Judge

Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboard with N rows and M columns.

Every day after work, Edward will place a chess piece on a random empty cell. A few days later, he found the chessboard was dominated by the chess pieces. that means there is at least one chess piece in every row. also, there is at least one chess piece in every column.

"That's interesting !" Edward said. He wants to know the expectation number of days to make an empty chessboard of N × M dominated. Please write a program to help him.

Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

There are only two integers N and M (1 <= N, M <= 50 ).

Output
For each test case, output the expectation number of days.

Any solution with a relative or absolute error of at most 10-8 will be accepted.

Sample Input
2
1 3
2 2

Sample Output
3.000000000000
2.666666666667

Analysis: This is the first question of the Mudanjiang field competition in 2014 =
We can follow the push probability, and finally E = p1 * n1 + p2 * n2 .....
You can also reverse the expectation.

I use dp [I] [j] [k] to indicate that there is an I row, and the j column is controlled on the k Day.
When the pawns are placed on the k + 1 day, there are four States to be transferred;
Dp [I] [j] [k + 1] * p1
Dp [I + 1] [j] [k + 1] * p2
Dp [I] [j + 1] [k + 1] * p3
Dp [I + 1] [j + 1] [k + 1] * p4
The probability here is just a line ~

# Include
  
   
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# Include
     
      
# Include const int N = 66*66; using namespace std; int n, m; double dp [55] [55] [N]; // dp [I] [j] [k] indicates that a row of I meets the condition that at least one piece is required in each row, and a column of j meets the condition that at least one piece is required in each column, A total of K lattice int main () {int T; scanf ("% d", & T); while (T --) {scanf ("% d ", & n, & m); int total = n * m; memset (dp, 0, sizeof (dp); dp [1] [1] [1] = 1.0; for (int I = 1; I <= n; I ++) for (int j = 1; j <= m; j ++) for (int k = max (I, j); k <= I * j; k ++) {if (I = n & j = m) break; // cout <