poj2151--Check the difficulty of problems（概率dp第四弹，复杂的计算）

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5009		Accepted: 2206

Description

Organizing a programming contest is not an easy job. To avoid making the problems too difficult, the organizer usually expect the contest result satisfy the following two terms: 
1. All of the teams solve at least one problem. 
2. The champion (One of those teams that solve the most problems) solves at least a certain number of problems. 

Now the organizer has studied out the contest problems, and through the result of preliminary contest, the organizer can estimate the probability that a certain team can successfully solve a certain problem. 

Given the number of contest problems M, the number of teams T, and the number of problems N that the organizer expect the champion solve at least. We also assume that team i solves problem j with the probability Pij (1 <= i <= T, 1<= j <= M). Well, can you calculate the probability that all of the teams solve at least one problem, and at the same time the champion team solves at least N problems? 

Input

The input consists of several test cases. The first line of each test case contains three integers M (0 < M <= 30), T (1 < T <= 1000) and N (0 < N <= M). Each of the following T lines contains M floating-point numbers in the range of [0,1]. In these T lines, the j-th number in the i-th line is just Pij. A test case of M = T = N = 0 indicates the end of input, and should not be processed.

Output

For each test case, please output the answer in a separate line. The result should be rounded to three digits after the decimal point.

Sample Input

2 2 2
0.9 0.9
1 0.9
0 0 0

Sample Output

0.972

题目：给出m个题，t个队伍，和每一个队伍做对每一个题的概率，问每一个队都做出题目，且有做对n或n以上题目的队的概率是多少？

转化。问题能够转化为：每一个队都做出1题或1题以上的概率 - 每一个队都做出1题到n-1题内的概率。

求每一个队做对k个题的概率。

dp[i][j][k]表示第i个队在前j个题目中做对k个的概率。

首先dp[i][0][0] = 1.0 ， 求解出dp[i][m][k]得到我们要求的概率

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
double dp[1005][32][32] ;
double p[1005][32] , p1 , p2 , temp ;

int main()
{
    int i , j , k , m , n , t ;
    while(scanf("%d %d %d", &m, &t, &n) && m+t+n != 0)
    {
        for(i = 1 ; i <= t ; i++)
            for(j = 1 ; j <= m ; j++)
                scanf("%lf", &p[i][j]);
        memset(dp,0,sizeof(dp));
        for(i = 1 ; i <= t ; i++)
        {
            dp[i][0][0] = 1.0 ;
            for(j = 1 ; j <= m ; j++)
            {
                for(k = 0 ; k <= j ; k++)
                {
                    if( k != 0 )
                        dp[i][j][k] += dp[i][j-1][k-1] * p[i][j] ;
                    if( k != j )
                        dp[i][j][k] += dp[i][j-1][k] * ( 1.0 - p[i][j] ) ;
                        //printf("%.2lf ", dp[i][j][k]) ;
                }
                //printf("
");
            }
            //printf("**
");
        }
        p1 = p2 = 1.0 ;
        for(i = 1 ; i <= t ; i++)
        {
            p1 *= ( 1.0 - dp[i][m][0] ) ;
            temp = 0.0 ;
            for(k = 1 ; k < n ; k++)
                temp += dp[i][m][k] ;
            p2 *= temp ;
        }
        printf("%.3lf
", p1-p2);
    }
    return 0;
}