POJ-2151 Check the difficulty of problems （DP）

Check the difficulty of problems

http://poj.org/problem?id=2151

Time Limit: 2000MS		Memory Limit: 65536K

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

题目大意：有t只队伍，m道题，求所有队伍都至少做出一题，并且存在队伍做出的题数大于等于n的概率？

可以用dp求得每支队伍做出k道题的概率

设dp[i][j][k]表示第i只队伍在前j题中做出k题的概率

则状态转移方程为：dp[i][j][k]=(1-p[i][j])*dp[i][j-1][k]+p[i][j]*dp[i][j-1][k-1];

总是想着正面求解，而忘了正难则反的原理

首先统计每个人至少做出一题的概率：prob_1=1-∏dp[i][m][0];

再统计每个人做出的题目数目均在1~n-1之间的概率：prob_2=∏(∑dp[i][m][k])

则满足题意的概率为：prob_1-prob_2

#include <cstdio>
#include <cmath>
#include <cstring>

using namespace std;

const int MAXN=35;
const double EPS=0.000001;

int m,t,n;
double dp[1005][MAXN][MAXN];//dp[i][j][k]表示第i只队伍在前j题中做出k题的概率
double p[1005][MAXN],prob_1,prob_2;

int main() {
    while(scanf("%d%d%d",&m,&t,&n),m!=0||t!=0||n!=0) {
        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][0][0]=1;
            for(int j=1;j<=m;++j) {
                dp[i][j][0]=(1-p[i][j])*dp[i][j-1][0];
            }//初始化第i支队伍的概率

            for(int j=1;j<=m;++j) {
                dp[i][j-1][j]=0;
                for(int k=1;k<=j;++k) {
                    dp[i][j][k]=(1-p[i][j])*dp[i][j-1][k]+p[i][j]*dp[i][j-1][k-1];
                }
            }
        }

        prob_1=prob_2=1;
        for(int i=1;i<=t;++i) {
            prob_1*=(1-dp[i][m][0]);
            double sum=0;
            for(int j=1;j<n;++j) {//统计第i队做出1~n-1道题的概率和
                sum+=dp[i][m][j];
            }
            prob_2*=sum;
        }
        printf("%.3lf\n",prob_1-prob_2);
    }
    return 0;
}