poj 2151 Check the difficulty of problems

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本文介绍了一道关于ACM竞赛中计算各队伍解题概率的问题，通过动态规划求解每支队伍至少解出一题及冠军队解出指定数量题目的概率。

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题目链接：点击打开链接

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

Sample Output

0.972

题意：
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

<span style="font-size:24px;">#include <iostream>
#include<cstdio>
using namespace std;


double dp[1111][50][50];
double p[1111][50];
double s[1111][50];

int main()
{
    int m,t,n;
    while(cin>>m>>t>>n)
    {
        if(m==0&&t==0&&n==0)break;
        for(int i=1;i<=t;i++)
        {
            for(int j=1;j<=m;j++)
            {
                scanf("%lf",&p[i][j]);
            }
        }
        for(int i=1;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 j=1;j<=m;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]);
                }
            }
             s[i][0]=dp[i][m][0];
             for(int k=1;k<=m;k++)s[i][k]=s[i][k-1]+dp[i][m][k];
        }
        double p1=1;
        double p2=1;
        for(int j=1;j<=t;j++)
        {
            p1=p1*(1-s[j][0]);
            p2=p2*(s[j][n-1]-s[j][0]);
        }
        printf("%.3f\n",p1-p2);
    }
    return 0;
}</span>