POJ2151——Check the difficulty of problems

Check the difficulty of problems

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 6784 Accepted: 2945

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

Source

POJ Monthly,鲁小石

题意：

组织一次程序设计大赛可不是什么容易的事，为了避免题目太难，一般都遵循以下规则：

1.所有的团队都能做出一题。（传说中的签到题）

2.冠军队（做对题数量最多的，没有罚时吗？）做对题数量不会太少。（就是不会出现大家只做对1，2道题，冠军队只比其他人多对1，2道题）

现在组织者已经给出了比赛的题目，并通过初步比赛的结果，组织者可以估计团队可以成功地解决问题的概率。你能计算出在所有队伍都至少解出一道，且冠军队伍解出n（0<n<=m）道题的概率吗？

思路：

这题不是哈希，是概率dp。

首先我们设dp[i][j][k]为第 i 组前 j 个题做对 k 道题目的概率。用另一个数组p[i][j]存下第 i 组做对第 j 题的概率。

递推式就有：dp[i][j][k] = dp[i][j-1][k-1]*(p[i][j])+dp[i][j-1][k]*(1-p[i][j])

最后所有团队都至少做对1题的概率减去所有团队都至少做对1～n-1道题的概率就完了。

用%f输出。用%f输出。用%f输出。

#include <iostream>#include <cstdlib>#include <cstdio>using namespace std;int main(){    int m, t, n;    double p[1001][31];    double dp[1001][31][31];    //第 i 组前 j 个题做对 k 道题目的概率    double tdp[1001][31];      //第 i 组解出不多于 j 题的概率    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++ )            {                cin>>p[i][j];            }            dp[i][0][0] = 1.0;            dp[i][0][1] = 0.0;            for( int j = 1;j <= m;j++ )                dp[i][j][0] = dp[i][j-1][0]*(1.0-p[i][j]);            for( int j = 1;j <= m;j++ )            {                for( int k = 1;k <= m;k++ )                {                    dp[i][j][k] = dp[i][j-1][k]*(1.0-p[i][j]) + dp[i][j-1][k-1]*p[i][j];                }            }            tdp[i][0] = dp[i][m][0];            for( int j = 1;j <= m;j++ )            {                tdp[i][j] = tdp[i][j-1] + dp[i][m][j];            }        }        double p1 = 1.0;      //至少做对一道题目的概率        double p2 = 1.0;     //至少做出1~n-1道题目的概率        for( int i = 1;i <= t;i++ )        {            p1 *= (tdp[i][n-1] - tdp[i][0]);            p2 *= (1.0 - tdp[i][0]);        }        printf("%.3f\n",p2-p1);    }    return 0;}