POJ2151: Check the difficulty of problems (probability DP)

Description

Organizing a programming contest is not an easy job. To avoid making the problems too difficult, the organizer usually failed CT 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 has CT 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 shoshould not be processed.
Output

For each test case, please output the answer in a separate line. The result shocould 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 Nima, the question card is accurate. The output must be single-precision but not dual-precision. Otherwise, WA will die. Question: There are t-teams, m-questions. The champion should have at least n questions. The question is the probability that each team will have at least one question and the champion will have at least n questions: there is really not much left in high school knowledge. I will repost the explanations from other blogs in detail below. Basically, I can look at this idea and turn it into code. Pay attention to accuracy.

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int n,m,t;double dp[1005][35][35],s[1005][35],p[1005][35];int main(){    int i,j,k;    double p1,p2;    while(~scanf("%d%d%d",&m,&t,&n))    {        if(!m && !t && !n)            break;        for(i = 1; i<=t; i++)            for(j = 1; j<=m; j++)                scanf("%lf",&p[i][j]);        memset(dp,0,sizeof(dp));        memset(s,0,sizeof(s));        for(i = 1; i<=t; i++)        {            dp[i][0][0] = 1.0;            for(j = 1; j<=m; j++)                dp[i][j][0] = dp[i][j-1][0]*(1-p[i][j]);            for(j = 1; j<=m; j++)                for(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][0] = dp[i][m][0];            for(k = 1; k<=m; k++)                s[i][k] = s[i][k-1]+dp[i][m][k];        }        p1 = p2 = 1.0;        for(i = 1; i<=t; i++)            p1*=(s[i][m]-s[i][0]);        for(i = 1; i<=t; i++)            p2*=(s[i][n-1]-s[i][0]);        printf("%.3f\n",p1-p2);    }    return 0;}