poj 2151 Check the difficulty of problems(機率DP)

標籤：dp-機率dp

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4680		Accepted: 2049

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 20.9 0.91 0.90 0 0

Sample Output

0.972

ans = (1-第1隊做出0題)*(1-第2隊做出0題)...(1-第n隊做出0題)-(第1隊做1-N題)*(第2隊做1-N題)...(第n隊做1-N題)。

藍色部分為條件1，紅色部分為條件2.

dp[t][m][i]表示第t隊在前m題中做出i題：

dp[t][m][i] =dp[t][m-1][i-1]*P[m][i]+dp[t][m-1][i]*(1-P[m][i]),i>0

dp[t][m][i] = dp[t][m-1][i]*(1-P[m][i]),i==0.

#include <iostream>#include <cstdio>#include <cmath>#include <string>#include <cstring>#include <cstdlib>using namespace std;const int maxt = 1010;const int maxm = 35;double dp[maxt][maxm][maxm] , P[maxt][maxm];int N , M , T;void initial(){for(int i = 0; i < maxt; i++){for(int j = 0; j < maxm; j++){P[i][j] = 0;for(int k = 0; k < maxm; k++) dp[i][j][k] = 0;}}}void readcase(){for(int i = 1; i <= T; i++){for(int j = 1; j <= M; j++) scanf("%lf" , &P[i][j]);}}void computing(){for(int i = 0; i <= T; i++) dp[i][0][0] = 1.0;for(int t = 1; t <= T; t++){for(int m = 1; m <= M; m++){for(int i = 0; i <= min(m , N); i++){if(i)dp[t][m][i] = dp[t][m-1][i-1]*P[t][m]+dp[t][m-1][i]*(1.0-P[t][m]);else dp[t][m][i] = dp[t][m-1][i]*(1.0-P[t][m]);}}}double ans = 1.0;for(int i = 1; i <= T; i++) ans = ans*(1.0-dp[i][M][0]);//cout << ans << endl;double tem = 1.0;for(int t = 1; t <= T; t++){double ttem = 0.0;for(int i = 1; i < N; i++) ttem += dp[t][M][i];tem *= ttem;}printf("%.3lf\n" , ans-tem);}int main(){while(scanf("%d%d%d", &M , &T , &N) && (N!=0 || M!=0 || T!=0)){initial();readcase();computing();}return 0;}

poj 2151 Check the difficulty of problems(機率DP)