POJ 2151 Check the difficulty of problems（概率dp）

题目链接：http://poj.org/problem?id=2151

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 20.9 0.91 0.90 0 0

Sample Output

0.972

题意：

比赛中，共 m 道题，t 个队，p[i][j]表示第 i 队解出第 j 题的概率

问:每队至少解出一题且冠军队至少解出 n 道题的概率。

PS：

p[i][j]表示第i个队伍做对第j题的概率。dp[i][j][k]表示第i个队伍在前j题中做对了k道的概率。
dp[i][j][k] = dp[i][j-1][k-1]*(p[i][j])+dp[i][j-1][k]*(1-p[i][j]);
再求出每个队都至少做对 1 道题的概率：tt *= 1 - dp[i][m][0];
最后再减去每个队都只做对了 1 ～ n-1 题的概率；
即：(把每个队做对 1 ～ n-1 题的概率相加后，并把每个队的结果相乘）;

代码如下：

//PS：p[i][j]表示第i个队伍做对第j题的概率。dp[i][j][k]表示第i个队伍在前j题中做对了k道的概率。////dp[i][j][k] = dp[i][j-1][k-1]*(p[i][j])+dp[i][j-1][k]*(1-p[i][j]);////再求出每个队都至少做对 1 道题的概率：tt *= 1 - dp[i][m][0];////最后再减去每个队都只做对了 1 ～ n-1 题的概率；//即：(把每个队做对 1 ～ n-1 题的概率相加后，并把每个队的结果相乘）#include <cstdio>#include <cstring>double p[1017][47];double dp[1017][47][47];int main(){    int n, m, t;    while(~scanf("%d%d%d",&m,&t,&n))    {        if(m==0 && t==0 && n==0)            break;        memset(p, 0,sizeof(p));        memset(dp, 0,sizeof(dp));        for(int i = 0; i < t; i++)        {            for(int j = 1; j <= m; j++)            {                scanf("%lf",&p[i][j]);            }        }        for(int i = 0; i < t; i++)        {            dp[i][0][0] = 1;            for(int j = 1; j <= m; j++)            {                dp[i][j][0] += dp[i][j-1][0] * (1-p[i][j]);                for(int k = 1; k <= j; k++)                {                    dp[i][j][k] = dp[i][j-1][k-1]*p[i][j]+dp[i][j-1][k]*(1-p[i][j]);                }            }        }        double tt = 1;        for(int i = 0; i < t; i++)        {            tt*=(1 - dp[i][m][0]);        }        double temp = 1, t_sum;        for(int i = 0; i < t; i++)        {            t_sum = 0;            for(int j = 1; j <= n-1; j++)            {                t_sum+=dp[i][m][j];            }            temp *= t_sum;        }        double ans = tt-temp;        printf("%.3lf\n",ans);    }    return 0;}