## POJ 2151 Check the difficulty of problems（概率DP）

Check the difficulty of problems
 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 3191 Accepted: 1416

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

/*
POJ 2151

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

dp[i][j][k]=dp[i][j-1][k-1]*p[j][k]+dp[i][j-1][k]*(1-p[j][k]);

*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
using namespace std;

double dp;
double s;
double p;
int main()
{
int M,N,T;
while(scanf("%d%d%d",&M,&T,&N)!=EOF)
{
if(M==0&&T==0&&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]=1;
for(int j=1;j<=M;j++)dp[i][j]=dp[i][j-1]*(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]=dp[i][M];
for(int k=1;k<=M;k++)s[i][k]=s[i][k-1]+dp[i][M][k];
}
double P1=1;
double P2=1;
for(int i=1;i<=T;i++)
{
P1*=(1-s[i]);
P2*=(s[i][N-1]-s[i]);
}
printf("%.3lf\n",P1-P2);
}
return 0;
}

posted on 2012-10-03 23:53  kuangbin  阅读(1502)  评论(0编辑  收藏

• 随笔 - 940
• 文章 - 0
• 评论 - 595
• 引用 - 0
JAVASCRIPT: