poj 2151 概率DP

最新推荐文章于 2020-11-21 20:14:24 发布

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本文介绍了一种使用动态规划方法来计算编程比赛中所有队伍至少解决一个问题且冠军队伍解决的问题数量达到一定标准的概率。通过构建概率矩阵并运用递推公式，文章详细展示了如何计算这一复杂问题。

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题目连接：http://poj.org/problem?id=2151

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5173		Accepted: 2289

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

Source

POJ Monthly,鲁小石

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思路：这道题dp的过程很好想，但是概率问题不太好弄啊~我概率木有学好，左想右想才弄懂~

（1）

我们用 dp[i][j][k] 表示第 i个队在前j个题目中做出来k个题目的概率

那么转移方程可以表示为 dp[i][j][k]=dp[i][j-1][k-1]*p[i][j]+dp[i][j-1][k]*(1-p[i][j]);

要注意初始化~

（2）

然后就是概率问题了，用P1表示所有队出题数至少1道的概率，P2表示所有队出题数在1~n-1之间的概率;最后p1-p2即为所求

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;

int m,t,n;
double p[1100][55];
double dp[1100][55][55];
double s[1100][55];

int main()
{
  while(scanf("%d%d%d",&m,&t,&n)!=EOF)
  {
    if( m+t+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]=1.0;
      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 j=1;j<=m;j++)
        s[i][j]=s[i][j-1]+dp[i][m][j];
    }


    double p1=1.0,p2=1.0;
    for(int i=1;i<=t;i++)
    {
     p1*=(1-s[i][0]);
     p2*=(s[i][n-1]-s[i][0]);
    }
    printf("%.3lf\n",p1-p2);

  }
  return 0;
}