poj 2151 Check the difficulty of problems 概率dp

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3617		Accepted: 1580

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

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f[i][j][k]表示第i组队伍做前j道题有k道正确的概率。。。

显然有 f[i][j][k]=f[i][j-1][k]*(1-p[i][j])+f[i][j-1][k-1]*p[i][j];

至少有一组做出n道以上的题且每组至少做出1道题的概率等于每组至少做出1题的概率-没有一组做出n道题以上的概率。

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#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn=1111;

int m,t,n;

double p[maxn][44];
double f[maxn][44][44];

int main()
{
    while (~scanf("%d%d%d",&m,&t,&n))
    {
        if (m==0&&t==0&&n==0) break;
        memset(f,0,sizeof(f));
        memset(p,0,sizeof(p));
        for (int i=1;i<=t;i++)
        {
            for (int j=1;j<=m;j++)
            {
                scanf("%lf",&p[i][j]);
            }
        }
        for (int i=1;i<=t;i++)
        {
            f[i][0][0]=1;
            for (int j=1;j<=m;j++)
            {
                for (int k=0;k<=j;k++)
                {
                    if (k==0)
                    {
                        f[i][j][k]=f[i][j-1][k]*(1-p[i][j]);
                    }
                    else
                    {
                        f[i][j][k]=f[i][j-1][k]*(1-p[i][j])+f[i][j-1][k-1]*p[i][j];
                    }
                }
            }
        }
        double ans=1,sum=0,ans2=1;
        for (int i=1;i<=t;i++)
        {
            sum=0;
            for (int j=1;j<=m;j++)
            {
                sum+=f[i][m][j];
            }
            ans*=sum;
        }
        for (int i=1;i<=t;i++)
        {
            sum=0;
            for (int j=1;j<n;j++)
            {
                sum+=f[i][m][j];
            }
            ans2*=sum;
        }
        ans-=ans2;
        printf("%0.3lf\n",ans);
    }
    return 0;
}