poj-2151 Check the difficulty of problems 概率+DP_poj 2151 check the difficulty of problems-优快云博客

本文链接：https://blog.youkuaiyun.com/chAngE_AC/article/details/47720423

本文介绍了一个程序设计比赛背景下，计算所有队伍至少解决一个问题且冠军队伍解决至少N个问题的概率问题。通过概率+动态规划的方法，给出了详细的解题思路及代码实现。

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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：在程序设计比赛中有T个队，M道题，冠军最少要对N道题，输入i队完成第j道题的概率，最后得到所有队至少完成1道题且有冠军产生的概率。

思路：该题为概率+DP，解题关键就在于找到每队完成K题（1<=K<=M）的概率，即找到该状态转移方程。不多说废话，本人也是参考的其他人的推导方法。

下面是本人按照上图的思路写的代码，思路很好理解，一次AC。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#define INF 0x3f3f3f3f
#define mod 1000000007

using namespace std;

double dp[1005][31][31];
double p[1005][31],s[1005][31];
int m,t,n;
void Clear()
{
    int i,j;
    memset(s,0,sizeof(s));
    memset(dp,0,sizeof(dp));
    for (i=1;i<=t;i++)
    {
        dp[i][0][0]=1.0;
    }
    for (i=1;i<=t;i++)
    {
        for (j=1;j<=m;j++)
        {
            dp[i][j][0] = dp[i][j-1][0]*(1-p[i][j]);
        }
    }
}
int main()
{
    int i,j,k;
    while(scanf("%d%d%d",&m,&t,&n),n||m||t)
    {
        for (i=1;i<=t;i++)
        {
            for (j=1;j<=m;j++)
            {
                scanf("%lf",&p[i][j]);
            }
        }
        Clear();
        for (i=1;i<=t;i++)
        {
            for (j=1;j<=m;j++)
            {
                for (k=1;k<=m;k++)
                {
                    dp[i][j][k]=dp[i][j-1][k-1]*p[i][j]+dp[i][j-1][k]*(1-p[i][j]);
                }
            }
        }
        for (i=1;i<=t;i++)
        {
            for (k=1;k<=m;k++)
            {
                s[i][k]=s[i][k-1]+dp[i][m][k];
            }
        }
        double p1=1,p2=1;
        for (i=1;i<=t;i++)
        {
            p1 *= s[i][m]-s[i][0];
            p2 *= s[i][n-1]-s[i][0];
        }
        printf("%.3lf\n",p1-p2);
    }
    return 0;
}