poj2151（概率DP）

地址：http://poj.org/problem?id=2151

Check the difficulty of problems

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 4188 Accepted: 1849

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：t个人参加有m道题的比赛，问每个人至少做出一道题，冠军最少做出n道题的概率。

思路：本题要求的概率是P（所有人解出至少一题）-P（所有人做的题数在1~n-1之间）。

            这两个概率的求解要用到DP。

            转移方程：dp[j][k]=dp[j-1][k-1]*p[i][j]+dp[j-1][k]*(1-p[i][j])，dp[j][k]表示在j题中作对k道的概率，p[i][j]表示第i人作对j题的概率。

代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;double p[1010][31],dp[31][31];int main(){    int m,t,n;    while(scanf("%d%d%d",&m,&t,&n)>0,m|t|n)    {        for(int i=1;i<=t;i++)            for(int j=1;j<=m;j++)                scanf("%lf",&p[i][j]);        double ans=1,ans1=1;        for(int i=1;i<=t;i++)        {            double ret=0;            dp[0][0]=1;            for(int j=1;j<=m;j++)            {                dp[j][0]=dp[j-1][0]*(1-p[i][j]);                for(int k=1;k<j;k++)                    dp[j][k]=(dp[j-1][k-1]*p[i][j]+dp[j-1][k]*(1-p[i][j]));                dp[j][j]=dp[j-1][j-1]*p[i][j];            }            for(int j=1;j<n;j++)                ret+=dp[m][j];            ans1*=ret;            ans*=(1-dp[m][0]);        }        printf("%.3f\n",ans-ans1);  //ans是所有人解出至少一题，ans1是所有人做的题数在1~n-1之间    }    return 0;}