POJ2151:Check the difficulty of problems(概率DP)

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

题意+题解

ACM比赛中，共M道题，T个队，pij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。

解析：DP
设dp[i][j][k]表示第i个队在前j道题中解出k道的概率
则：
dp[i][j][k]=dp[i][j-1][k-1]*p[j][k]+dp[i][j-1][k]*(1-p[j][k]);
先初始化算出dp[i][0][0]和dp[i][j][0];
设s[i][k]表示第i队做出的题小于等于k的概率
则s[i][k]=dp[i][M][0]+dp[i][M][1]+``````+dp[i][M][k];

则每个队至少做出一道题概率为P1=(1-s[1][0])*(1-s[2][0])*```(1-s[T][0]);
每个队做出的题数都在1~N-1的概率为P2=(s[1][N-1]-s[1][0])*(s[2][N-1]-s[2][0])*```(s[T][N-1]-s[T][0]);

最后的答案就是P1-P2

每个队至少做出一道题的概率-p2，就是减法原理，意思就是去除了全是1-（n-1）的情况，然后就是
赢的队至少解出了n道题。

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<iostream>
#define N 1007
#define M 57
using namespace std;

int m,n,t;
double f[N][M][M],z[N][M],gx[N][M];


int main()
{
    while(~scanf("%d%d%d",&m,&t,&n)&&(n+m+t))
    {
        for (int i=1;i<=t;i++)
            for (int j=1;j<=m;j++)
                scanf("%lf",&z[i][j]);
        memset(f,0,sizeof(f)),memset(gx,0,sizeof(gx));        
        for (int i=1;i<=t;i++)
        {
            f[i][0][0]=1;
            for (int j=1;j<=m;j++)
                f[i][j][0]=f[i][j-1][0]*(1-z[i][j]);
            for (int j=1;j<=m;j++)
                for (int k=1;k<=j;k++)
                    f[i][j][k]=f[i][j-1][k-1]*z[i][j]+f[i][j-1][k]*(1-z[i][j]);
            gx[i][0]=f[i][m][0];
            for (int j=1;j<=m;j++)
                gx[i][j]=gx[i][j-1]+f[i][m][j];    
        }
        double ans1,ans2;
        ans1=ans2=1;
        for (int i=1;i<=t;i++)
        {
            ans1=ans1*(1-gx[i][0]);
            ans2=ans2*(gx[i][n-1]-gx[i][0]);
        }
        printf("%.3f
",ans1-ans2);
    }
}