zoj 3822概率dp

zoj 3822概率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

分析： 這是2014年牡丹江現場賽一題==
可以順著推概率，最後E=p1*n1+p2*n2…..
也可以逆著推期望。

我用的是dp[i][j][k]表示有i行，j列在第k天被控制好了。
那麼當第k+1天放棋子的時候就有四種狀態轉移；
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
這裡的概率分別求就行~

#include 
#include 
#include 
#include 
#include 
const int N=66*66;
using namespace std;
int n,m;
double dp[55][55][N];
//dp[i][j][k]表示有i行滿足每行至少有一個棋子的條件，有j列滿足每列至少有一個棋子的條件，共占據了K個格子

int main()
{
   int T;
   scanf("%d",&T);
   while(T--)
   {
     scanf("%d%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<