poj 2151 -- Check the difficulty of problems

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4700		Accepted: 2057

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

思路： 概率dp。
　　　　dp[i][j][k] 表示第i个人前j个题做对k个的概率。
　　　　dp[i][j][k] = dp[i][j-1][k]*(1-g[i][j]) + dp[i][j][k-1]*g[i][j];

/*======================================================================
 *           Author :   kevin
 *         Filename :   CheckTheDifficultOfProblem.cpp
 *       Creat time :   2014-07-25 10:33
 *      Description :
========================================================================*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#define clr(a,b) memset(a,b,sizeof(a))
#define M 1005
using namespace std;
double g[M][35],dp[M][35][35];
int main(int argc,char *argv[])
{
    int n,t,m;
    while(scanf("%d%d%d",&n,&t,&m), n | t | m){
        for(int i = 0; i < t; i++){
            for(int j = 1; j <= n; j++){
                scanf("%lf",&g[i][j]);
            }
        }
        clr(dp,0);
        for(int i = 0; i < t; i++){
            dp[i][0][0] = 1;
            for(int j = 1; j <= n; j++){
                dp[i][j][0] = dp[i][j-1][0] * (1-g[i][j]);
                for(int k = 1; k <= j; k++){
                    dp[i][j][k] = dp[i][j-1][k]*(1-g[i][j])+dp[i][j-1][k-1]*g[i][j];
                }
            }
        }
        double s = 1;
        for(int i = 0; i < t; i++){
            s *= (1-dp[i][n][0]);
        }
        double ss = 1;
        for(int i = 0; i < t; i++){
            double tem = 0;
            for(int j = 1; j < m; j++){
                tem += dp[i][n][j];
            }
            ss *= tem;
        }
        s -= ss;
        printf("%.3lf\n",s);
    }
    return 0;
}