POJ 2151 Check the difficulty of problems

以前做过的题目了。。。。补集+DP

       Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4091		Accepted: 1811

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

Source

POJ Monthly,鲁小石 

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

using namespace std;

int M,T,N;
double a[1100][50][50],s[1100][50],p1,pn,solve[1100][50];

int main()
{
    while(~scanf("%d%d%d",&M,&T,&N))
    {
        if((M||T||N)==0) break;
        for(int i=1;i<=T;i++) for(int j=1;j<=M;j++) scanf("%lf",&solve[i][j]);
        memset(a,0,sizeof(a)); memset(s,0,sizeof(s));
        for(int i=1;i<=T;i++)
        {
            a[i][0][0]=1;
            for(int j=1;j<=M;j++)
            {
                a[i][j][0]=a[i][j-1][0]*(1-solve[i][j]);
            }
        }
        for(int i=1;i<=T;i++)
        {
            for(int j=1;j<=M;j++)
            {
                for(int k=1;k<=j;k++)
                {
                    a[i][j][k]=a[i][j-1][k-1]*solve[i][j]+a[i][j-1][k]*(1-solve[i][j]);
                }
            }
        }
        for(int i=1;i<=T;i++)
        {
            s[i][0]=a[i][M][0];
            for(int j=1;j<=M;j++)
            {
                s[i][j]=s[i][j-1]+a[i][M][j];
            }
        }
        p1=pn=1.;
        for(int i=1;i<=T;i++)
        {
            p1*=s[i][M]-s[i][0];
            pn*=s[i][N-1]-s[i][0];
        }
        printf("%.3lf
",p1-pn);
    }
    return 0;
}