poj 2151 Check the difficulty of problems（概率dp）

题目链接

Check the difficulty of problems

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 5482 Accepted: 2414

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：M道题目，T个人，已经每个人做出每道题目的概率，求所有人都过题，但是最多过N到题的概率。

题解：首先用dp[i][j][k]表示第i个人对于前j个题目，做出来k道题目的概率。通过这个dp我们先求出第i个人过j道题目的概率。然后再用f[i][j]表示前i个人，都过了题，最多过了j道题的概率。最后统计答案即可。

代码如下：

#include<stdio.h>#include<iostream>#include<algorithm>#include<set>#include<vector>#include<string.h>#include<map>#define inff 0x3fffffff#define nn 1100#define mod 1000003typedef __int64 LL;typedef unsigned __int64 LLU;using namespace std;int m,t,n;double p[nn][33];double dp[nn][33][33];double sum[nn][33];double f[nn][33];int main(){    int i,j,k;    while(scanf("%d%d%d",&m,&t,&n)&&(m+t+n))    {        for(i=1;i<=t;i++)        {            for(j=1;j<=m;j++)            {                scanf("%lf",&p[i][j]);            }        }        memset(dp,0,sizeof(dp));        for(i=1;i<=t;i++)        {            dp[i][0][0]=1.0;            for(j=1;j<=m;j++)            {                for(k=0;k<=j;k++)                {                    dp[i][j][k]+=dp[i][j-1][k]*(1-p[i][j]);                    if(k>0)                        dp[i][j][k]+=dp[i][j-1][k-1]*p[i][j];                }            }        }        for(i=1;i<=t;i++)        {            sum[i][0]=dp[i][m][0];            for(j=1;j<=m;j++)            {                sum[i][j]=sum[i][j-1]+dp[i][m][j];            }        }        memset(f,0,sizeof(f));        f[0][0]=1.0;        for(i=1;i<=t;i++)        {            for(j=1;j<=m;j++)            {                f[i][j]+=f[i-1][j]*(sum[i][j]-sum[i][0]);                for(k=0;k<j;k++)                {                    f[i][j]+=f[i-1][k]*(sum[i][j]-sum[i][j-1]);                }            }        }        double ans=0;        for(i=n;i<=m;i++)        {            ans+=f[t][i];        }        printf("%.3lf\n",ans);    }    return 0;}