（POJ 2151）Check the difficulty of problems（概率DP+容斥）

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



比较简单的一道概率dp，然而还是没能自己想出来（现在一丢丢罪恶感都没。。。）
要求的概率=P（所有队伍都不挂机）-P（所有队伍的过题数都在[1,n-1]之间）
在DP的时候开一个三维数组：dp[i][j][k]表示第i支队伍在前j题中通过了k道题，这样在最后可以看dp[i][m][k]来得到每个队伍过某题数的概率。递推方程很好得出：dp[i][j][k]=dp[i][j-1][k]*(1-p[i][j])+dp[i][j-1][k-1]*p[i][j];

注意：①k=0时是特殊情况，dp值要单独计算！
　　　②提交时看好自己的printf格式，如果是G++需要将double型用%.3f输出，C++用%lf输出。

#include<iostream>
#include<cstdio>
#include<vector>
#include<set>
#include<map>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#define LL long long
#define mod 1000000007
#define inf 0x3f3f3f3f

using namespace std;


double a[1100][55];
double dp[1100][55][55];
int main()
{
    int m,t,n;
    while(~scanf("%d%d%d",&m,&t,&n))
    {
        if(t==0&&m==0&&n==0)
            break;
        memset(a,0,sizeof(a));
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=t;i++)
            for(int j=1;j<=m;j++)
                scanf("%lf",&a[i][j]);

        for(int i=1;i<=t;i++)
        {
            dp[i][1][0]=1-a[i][1];
            dp[i][1][1]=a[i][1];
        }
        for(int i=1;i<=t;i++)
            for(int j=2;j<=m;j++)
                for(int k=0;k<=j;k++)
                {
                    if(k==0)
                        dp[i][j][k]=dp[i][j-1][k]*(1-a[i][j]);
                    else
                        dp[i][j][k]=dp[i][j-1][k-1]*a[i][j]+dp[i][j-1][k]*(1-a[i][j]);
                }
        double sum=1;
        for(int i=1;i<=t;i++)
            sum*=(1-dp[i][m][0]);
        double ans=1;
        for(int i=1;i<=t;i++)
        {
            double tem=0;
            for(int j=1;j<=n-1;j++)
                tem+=dp[i][m][j];
            ans*=tem;
        }

        printf("%.3lf\n",sum-ans);
    }
    return 0;
}