ACM learning process -- zoj 3822 domination (question D of 2014 Mudanjiang semi-finals) (probability, mathematical recursion)

Description

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 chessboardNRows andMColumns.

Every day after work, Edward will place a chess piece on a random empty cell. A few days later, he found the chessboard wasDominatedBy 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 chessboardN×MDominated. Please write a program to help him.

Input

There are multiple test cases. The first line of input contains an integerTIndicating the number of test cases. For each test case:

There are only two integersNAndM(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

21 32 2

Sample output

3.20.000000002.666666666667


In this question, we can set a probability function P (I, j, k) to represent the probability to overwrite column J in the I row with K stones. We can first calculate all P values, then, the probability multiplication day sum is used to obtain the expectation.
However, for the recurrence of probability, you can consider the following four situations:
1. By P (I-1, J-1, k-1) in the row not covered cut its column not covered in the grid put a stone;
2, by P (I-1, J, k-1) put a stone in a row not covered;
3. Put a stone in a column not covered by P (I, J-1, k-1;
4, by P (I-1, J-1, k-1) in the covered rows and put a column in the grid stone;
However, note that when calculating P (n, m ,?) You do not need to add the fourth item.
However, each type has its own probability coefficient, which is embodied in the Code. (Number of grids meeting the condition/total number of remaining grids) PS: The key is to find the number of grids meeting the condition.


Code:

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <set>#include <map>#include <vector>#include <string>#include <queue>#define inf 0x3fffffff#define esp 1e-10#define N 55using namespace std;double p[N][N][N*N];double qt (int n, int m){    memset (p, 0, sizeof(p));    int len = n*m;    p[0][0][0] = 1;    for (int i = 1; i <= n; ++i)    {        for (int j = 1; j <= m; ++j)        {            for (int k = 1; k <= len; ++k)            {                p[i][j][k] = p[i-1][j-1][k-1]/(n*m-k+1)*(n-i+1)*(m-j+1);                p[i][j][k] += p[i-1][j][k-1]/(n*m-k+1)*(n-i+1)*j;                p[i][j][k] += p[i][j-1][k-1]/(n*m-k+1)*i*(m-j+1);                if (i != n || j != m)                    p[i][j][k] += p[i][j][k-1]/(n*m-k+1)*(i*j-k+1);            }        }    }    double ans = 0;    for (int k = 0; k <= len; ++k)        ans += p[n][m][k]*k;    return ans;}int main(){    //freopen ("test.txt", "r", stdin);    int T;    scanf ("%d", &T);    for (int times = 0; times < T; ++times)    {        int n, m;        scanf ("%d%d", &n, &m);        printf ("%.10lf\n", qt (n, m));    }    return 0;}

 

ACM learning process -- zoj 3822 domination (question D of 2014 Mudanjiang semi-finals) (probability, mathematical recursion)