poj2151

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 1790		Accepted: 788

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

2 2 2
0.9 0.9
1 0.9
0 0 0

Sample Output

0.972

Source

POJ Monthly,鲁小石

DP还不是很熟练，参考了 http://blog.sina.com.cn/s/blog_6635898a0100itre.html

题意：举办一次ACM竞赛，需要考虑两方面，一是每个队至少都能做出1道题目，二是冠军至少能做出n道题目。现在已知有m道题目，t支队伍，和n的值，以及每支队伍做出每道题目的概率pro[i][j]，求出这次比赛能保证上面两方面都会达到的概率。

 

思路：很好的概率题，用DP做。首先把问题的解可以转化为：每队均至少做一题的概率P1，减去每队做题数均在1到n-1之间的概率P2。P1不难求，P2比较难点，一开始以为用深搜，但觉得肯定是超时了。所以P2得用DP去做，设dp[i][j]表示在前i道题中共解出j道题的概率，则状态方程为：

      dp[i][0] = dp[i-1][0] * (1 - pro[k][i]);
      dp[i][j] = dp[i-1][j] * (1 - pro[k][i]) + dp[i-1][j-1] * pro[k][i]; (0 < j < n)
      dp[i][i] = dp[i-1][i-1] * pro[k][i];

考虑了边界情况，初始化可以简单很多，运行效率也提高了一点。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

float p[1001][31];
float dp[31][31];

int main()
{
    int m,t,n;
    float p1, pn, pi;

    while(cin>>m>>t>>n && (m+t+n))
    {
        memset(p,0,sizeof(p));
        memset(dp,0,sizeof(dp));

        for(int i=1; i<=t; i++)
        {
            for(int j=1; j<=m; j++)
            {
                cin>>p[i][j]; //the possibility of team i solve problem j;
            }
        }

        //compute p1 >=1
        p1=1;
        for(int i=1; i<=t; i++)
        {
            pi=1;
            for(int j=1; j<=m; j++)
            {
                pi*=(1-p[i][j]);
            }
            p1*=(1-pi);
        }

        //compute pn
        pn=1;
        for(int i=1; i<=t; i++)
        {
            dp[0][0]=1;
            for(int j=1; j<=m; j++)
            {
                dp[j][0]=dp[j-1][0]*(1-p[i][j]);
                for(int k=1; k<j; k++)
                {
                    dp[j][k]=dp[j-1][k]*(1-p[i][j])+dp[j-1][k-1]*p[i][j];
                }
                dp[j][j]=dp[j-1][j-1]*p[i][j];
            }
            pi=0;
            for(int j=1; j<n; j++)
            {
                pi+=dp[m][j];
            }
            pn*=pi;
        }

        printf("%.3f\n", p1 - pn);
    }
    return 0;
}