Check the difficulty of problems - POJ 2151 概率dp-优快云博客

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/40746533

本文探讨了在组织编程竞赛时如何通过概率计算确保每个团队都能解决至少一道问题，同时冠军团队能解决至少指定数量的问题。通过给定的每个团队解决每道题目概率，文章详细解释了计算该概率的方法。

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Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5037		Accepted: 2226

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：给你每个队做出来每一道题的概率，问使得每个队都至少做出一道题，并且第一的最做出来不少于N道题的概率。

思路：额额额……我这个思路比较笨，1-P（每个人都是1到n-1的概率）-P（有0出现的情况）。或者可以这么写

P（每个队都至少做出1题）-P（每个人都是1到n-1的概率）。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
double DP[1010][35],dp[2][35];
int n,m,p;
int main()
{
    int i,j,k,a,b;
    double f,ans,ret;
    while(~scanf("%d%d%d",&n,&m,&p) && n+m+p!=0)
    {
        for(i=1;i<=m;i++)
        {
            memset(dp,0,sizeof(dp));
            dp[0][0]=1;
            for(j=1;j<=n;j++)
            {
                if(j&1)
                  a=0,b=1;
                else
                  a=1,b=0;
                scanf("%lf",&f);
                for(k=j;k>=1;k--)
                   dp[b][k]=dp[a][k-1]*f+dp[a][k]*(1-f);
                dp[b][0]=dp[a][0]*(1-f);
            }
            for(j=0;j<=n;j++)
               DP[i][j]=dp[b][j];
        }
        ans=1;
        f=1;
        for(i=1;i<=m;i++)
        {
            ret=0;
            for(j=1;j<p;j++)
               ret+=DP[i][j];
            f*=ret;
        }
        ans-=f;
        f=1;
        for(i=1;i<=m;i++)
            f*=(1-DP[i][0]);
        ans-=1-f;
        printf("%.3f\n",ans);
    }
}