POJ 2151 Check the difficulty of problems

以前做过的题目了。。。。补集+DP

       Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4091		Accepted: 1811

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 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 int M,T,N;
 8 double a[1100][50][50],s[1100][50],p1,pn,solve[1100][50];
 9 
10 int main()
11 {
12     while(~scanf("%d%d%d",&M,&T,&N))
13     {
14         if((M||T||N)==0) break;
15         for(int i=1;i<=T;i++) for(int j=1;j<=M;j++) scanf("%lf",&solve[i][j]);
16         memset(a,0,sizeof(a)); memset(s,0,sizeof(s));
17         for(int i=1;i<=T;i++)
18         {
19             a[i][0][0]=1;
20             for(int j=1;j<=M;j++)
21             {
22                 a[i][j][0]=a[i][j-1][0]*(1-solve[i][j]);
23             }
24         }
25         for(int i=1;i<=T;i++)
26         {
27             for(int j=1;j<=M;j++)
28             {
29                 for(int k=1;k<=j;k++)
30                 {
31                     a[i][j][k]=a[i][j-1][k-1]*solve[i][j]+a[i][j-1][k]*(1-solve[i][j]);
32                 }
33             }
34         }
35         for(int i=1;i<=T;i++)
36         {
37             s[i][0]=a[i][M][0];
38             for(int j=1;j<=M;j++)
39             {
40                 s[i][j]=s[i][j-1]+a[i][M][j];
41             }
42         }
43         p1=pn=1.;
44         for(int i=1;i<=T;i++)
45         {
46             p1*=s[i][M]-s[i][0];
47             pn*=s[i][N-1]-s[i][0];
48         }
49         printf("%.3lf\n",p1-pn);
50     }
51     return 0;
52 }