POJ2151-Check the difficulty of problems

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Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8583		Accepted: 3656

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：给出几个队，几道题，第一名至少A的题数N，求每队都至少有一道A并且冠军队伍达到N题的概率

 

思路：只知道是概率dp（神犇说是简单的dp）主要还是怎么规划好状态。（初始化和状态转移方程）

 

AC代码：

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

int M, T, N;
double dp[1010][50][50];//dp[i][j][k] 第i队前j题过k道的概率 
double s[1010][50];//s[i][k] 第i队过小于等于k的概率 
double p[1010][50];

int main() {
	while(~scanf("%d %d %d", &M, &T, &N)) {
		if(M+T+N == 0)
			break;
		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;//前0题过0道概率为1 
			for(int j = 1; j <= M; j++) {
				dp[i][j][0] = dp[i][j-1][0] * (1-p[i][j]);//前j题过0道 = 前j-1题概率 * 本题不过概率 
			}
			for(int j = 1; j <= M; j++) { //前j题过k道 = 前j-1题过k-1道的概率 * 本题过的概率 + 前j-1题过k道 * 本题不过概率 
				for(int k = 1; k <= j; k++) {  //写成了k <= M     
					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];//过小于等于0道概率 就是前M题过0道概率 
			for(int k = 1; k <= M; k++) {
				s[i][k] = s[i][k-1] + dp[i][M][k];// s[i][k] = dp[i][M][0] + dp[i][M][1] + ...+ dp[i][M][k]; 
			}
		}
		double P1 = 1;
		double P2 = 1;
		for(int i = 1; i <= T; i++) {
			P1 *= (1-s[i][0]); // 所有队都至少过一道概率 
			P2 *= (s[i][N-1] - s[i][0]); //所有对都过 1~N-1道 概率 
		}
		printf("%.3lf\n", P1-P2); 
	}
}