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poj 2151

2024-07-17 20:41:01   220人阅读

  http://poj.org/problem?id=2151   
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Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 4873 Accepted: 2131

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 20.9 0.91 0.90 0 0

Sample Output

0.972
分析：求保证每个队至少做对一题，冠军队做对n个题的概率。
保证每个队至少做对一题，冠军队做对n个题的概率=每个队至少做对一道题-没有一个队做到n到题。（每个队最多做了n-1个题），
dp[i][j][k]表示第i个对做到j题，目前做对了k题。
dp[i][j[k]=dp[i][j-1][k]*(1-a[i][j])+dp[i][j-1][k-1]*a[i][j];
s[i][k]表示i对至少做对了k题的概率
注意边界。
#include<iostream>#include<cstdio>#include<cstring>using namespace std;double dp[1005][40][40];int main(){   int m,t,n,i,j,k;   double a[1005][40],cnt,ans,sum,s[1005][40];   while(~scanf("%d%d%d",&m,&t,&n))   {       memset(dp,0,sizeof(dp));       memset(s,0,sizeof(s));               cnt=1;               ans=1;               sum=1;       if(m==0&&t==0&&n==0)              break;        for(i=1;i<=t;i++)          {              for(j=1;j<=m;j++)              {                 scanf("%lf",&a[i][j]) ;                 cnt*=(1-a[i][j]);              }              ans*=(1-cnt);              cnt=1;          }           for(i=1;i<=t;i++)          {             dp[i][1][0]=1-a[i][1];              dp[i][1][1]=a[i][1];             for(j=2;j<=m;j++)                 dp[i][j][0]=dp[i][j-1][0]*(1-a[i][j]);             for(j=2;j<=m;j++)              {                 for(k=1;k<=j;k++)                 {                 dp[i][j][k]=dp[i][j-1][k]*(1-a[i][j])+dp[i][j-1][k-1]*a[i][j];                 }              }              for(k=1;k<=n-1;k++)                 s[i][n-1]+=dp[i][m][k];             }          for(i=1;i<=t;i++)          {             sum*=s[i][n-1];          }      printf("%.3lf\n",ans-sum);   }   return 0;}

 


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