【POJ2151】Check the difficulty of problems

最新推荐文章于 2021-06-18 18:27:09 发布

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最新推荐文章于 2021-06-18 18:27:09 发布

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分类专栏：概率DP poj 文章标签：概率DP POJ

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本文介绍了一个竞赛组织者如何通过预估各队伍解决特定编程问题的概率来计算所有队伍至少解决一个问题且冠军队伍至少解决指定数量问题的概率的方法。文章提供了一种动态规划的解决方案，并给出了完整的代码实现。

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8758		Accepted: 3720

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

解析：

令f[ i ][ j ][ k ]为到第 i 队前 j 道题完成 k 道的概率，于是易得：

f[ i ][ j ][ k ]=f[ i ][ j - 1 ][ k - 1 ]*p[ i ][ j ]+f[ i ][ j - 1 ][ k ]*(1.0 - p[ i ][ j ])。

代码：

//poj2151
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
using namespace std;

const int Max=1010;
int n,m,k;
double f[Max][35][35],ans,p[Max][35],sum;

int main()
{
	while(1)
	{
	  scanf("%d%d%d",&m,&n,&k);
	  if((!n)&&(!m)&&(!k)) break;
	  ans=sum=1,memset(f,0,sizeof(f));
	  for(int i=1;i<=n;i++) f[i][0][0]=1;
	  for(int i=1;i<=n;i++)
	    for(int j=1;j<=m;j++)
	      scanf("%lf",&p[i][j]);
	  for(int i=1;i<=n;i++)
	    for(int j=1;j<=m;j++)
	      for(int k=0;k<=j;k++)
	        f[i][j][k]=f[i][j-1][k-1]*p[i][j]+f[i][j-1][k]*(1.0-p[i][j]);
	  for(int i=1;i<=n;i++)
	  {
	  	double x=0,y=0; 
	  	for(int j=1;j<=k-1;j++) x+=f[i][m][j];
	  	for(int j=1;j<=m;j++) y+=f[i][m][j];
	  	ans*=y,sum*=x; 
	  }
	  printf("%.3f\n",ans-sum);
	}
	return 0;
}