Poj 2151 Check the difficulty of problems （概率DP）

最新推荐文章于 2019-07-17 15:56:41 发布

Grit_ICPC

最新推荐文章于 2019-07-17 15:56:41 发布

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分类专栏：概率DP 文章标签：概率DP POJ

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本文深入探讨了在编程竞赛中如何合理安排问题难度，确保每支队伍至少解决一道题，同时冠军队伍能够解决指定数量的问题。通过概率DP算法，计算出所有队伍至少解决一题且冠军队至少解决特定数量问题的概率。

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Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5779		Accepted: 2518

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

看完题目后如果知道概率DP的话，应该都能想到。关键是怎么用。

题意：
ACM比赛中，共M道题，T个队，pij表示第i队解出第j题的概率
问 每队至少解出一题且冠军队至少解出N道题的概率。

思路：

用dp[i][j][k]表示，第i个队，在前j题，解出k道题的概率。

s[i][k]代表第i个队解出题<=k个的概率

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
double p[1100][40],dp[1100][40][40],pro[1100][40];
int n,m,t;
int main()
{
    int i,j,k;
    while(scanf("%d%d%d",&m,&t,&n)!=EOF&&n&&m&&t)
    {
        memset(dp,0,sizeof(dp));
        memset(pro,0,sizeof(pro));
        for(i=1;i<=t;i++)
        {
            for(j=1;j<=m;j++)
            {
                scanf("%lf",&p[i][j]);//第i个人解出第j个题的概率，（输入最好，以1为下标开始，以防数组跃界，并容易选择基）
            }
        }
        for(i=1;i<=t;i++)
        {
            dp[i][0][0]=1;//题目数为1-m，所以解出第零个题的概率一定1,（以此作为加的基数）
            for(j=1;j<=m;j++)
            dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]); //先推i个队，在前j道题中做出0题的概率，以此做基推后边
            for(j=1;j<=m;j++)
            {
                for(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]);//当前的状态和前面的状态及现在能做出题目的概率相关
                }
            }
            pro[i][0]=dp[i][m][0];//第i队做出0<=题的概率，和在m个题中没做出题目的概率相同。
            for(k=1;k<=m;k++)
              pro[i][k]=pro[i][k-1]+dp[i][m][k];//当前做出题目<=k道的概率一定比在m道中做出k道题的概率大，肯定是由上一中情况推来的加上
        }
        double an1=1,an2=1;
        for(i=1;i<=t;i++)
        {
            an1*=(1-pro[i][0]);//此是各队做出题数>=1到的总概率
            an2*=(pro[i][n-1]-pro[i][0]);//每个队做出的题目个数在1~n-1的总概率<span style="white-space:pre">	</span>
        }
        printf("%.3f\n",an1-an2);
    }
    return 0;
}