poj 2151 Check the difficulty of problems

Check the difficulty of problems

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 6679 Accepted: 2905

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

提示

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，说到dp可是我的弱项，我也就只有去百度了。

首先我们设dp[i][j][k]为第 i 组做了 j 题成功AC k 题的概率。用另一个数组f[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]*(f[i][j])+dp[i][j-1][k]*(1-f[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。

示例程序

Source CodeProblem: 2151Code Length: 1085BMemory: 8176KTime: 47MSLanguage: GCCResult: AcceptedSource Code#include <stdio.h>#include <string.h>double dp[1000][31][31],f[1000][31];int main(){    int n,m,t,i,i1,i2;    double p1,p2,sum;    scanf("%d %d %d",&m,&t,&n);    while(n!=0||t!=0||m!=0)    {        p1=1;        p2=1;        for(i=0;t>i;i++)        {            for(i1=1;m>=i1;i1++)            {                scanf("%lf",&f[i][i1]);            }        }        for(i=0;t>i;i++)        {            dp[i][0][0]=1;            for(i1=1;m>=i1;i1++)            {                dp[i][i1][0]=dp[i][i1-1][0]*(1-f[i][i1]);                for(i2=1;i1>=i2;i2++)                {                    dp[i][i1][i2]=dp[i][i1-1][i2-1]*f[i][i1]+dp[i][i1-1][i2]*(1-f[i][i1]);                }            }        }        for(i=0;t>i;i++)        {            p1=p1*(1-dp[i][m][0]);        }        for(i=0;t>i;i++)        {            sum=0;            for(i1=1;n>i1;i1++)            {                sum=sum+dp[i][m][i1];            }            p2=p2*sum;        }        printf("%.3f\n",p1-p2);        scanf("%d %d %d",&m,&t,&n);    }    return 0;}