POJ 2151, Check the difficulty of problems

Time Limit: 2000MS  Memory Limit: 65536K
Total Submissions: 1015  Accepted: 391

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

---

//  POJ2151.cpp : Defines the entry point for the console application.
//

#include  < iostream >
#include  < iomanip >
using   namespace  std;

int  main( int  argc,  char *  argv[])
{
     int  N,M,T;
     double  TM[ 1001 ][ 31 ];

     double  DP[ 31 ][ 31 ];
     while  (cin  >>  M  >>  T  >>  N  &&  M  !=   0   &&  T  !=   0   &&  N  !=   0 )
    {
        memset(TM,  0 ,  sizeof (TM));
         for  ( int  i  =   0 ; i  <  T;  ++ i)
             for  ( int  j  =   0 ; j  <  M;  ++ j)
                scanf( " %lf " ,  & TM[i][j]);

         double  P1  =   1 ,P2  =   1 ;
         for  ( int  k  =   0 ; k  <  T;  ++ k)
        {
            memset(DP,  0 ,  sizeof (DP));
            DP[ 0 ][ 0 ]  =   1 ;
             for  ( int  i  =   1 ; i  <=  M;  ++ i)
                 for  ( int  j  =   0 ; j  <=  M;  ++ j)
                    DP[i][j]  =  (j  ==   0 ) ?  DP[i - 1 ][j]  *  ( 1   -  TM[k][i  -   1 ]):DP[i - 1 ][j - 1 ]  *  TM[k][i  -   1 ]  +  DP[i - 1 ][j]  *  ( 1   -  TM[k][i  -   1 ]);

            P1  *=  ( 1   -  DP[M][ 0 ]);

             double  P  =   0 ;
             for  ( int  i  =   1 ; i  <  N;  ++ i) P  +=  DP[M][i];

            P2  *=  P;
        }
        cout  <<   fixed   <<  showpoint  <<  setprecision( 3 ) <<  P1  -  P2  <<  endl;
    };
     return   0 ;
}