POJ2151:Check the difficulty of problems概率DP
Posted wans-caesar-02111007
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Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 8903 | Accepted: 3772 |
Description
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
Output
Sample Input
2 2 2
0.9 0.9
1 0.9
0 0 0
Sample Output
0.972
Source
Solution
概率DP
定义$dp[i][j][k]$表示第$i$个队伍做了$j$道题对了$k$道的概率。转移方程显然。
再定义$s[i][j]$表示第$i$个队对了不超过$j$道题的概率,$s[i][j]=sum{dp[i][t][k]},0<=k<=j$
然后所有队伍都对了至少一道题的概率就是$p1=prod{(1-s[i][0])}$
因为冠军队至少对了$n$道,那么最后的概率应该是$p1$减去每个队伍都没有达到$n$的概率:$prod{s[i][n-1]}$,在这里要注意,还要保证每个队伍都至少对了1道题,因为是递推转移得到$s$,不能把$0$的贡献算进去,所以$s$从2开始递推。
Code
#include<iostream> #include<cstdio> using namespace std; double dp[1005][35][35], s[1005][35], p[1005][35]; int m, t, n; int main() { while(scanf("%d%d%d", &m, &t, &n) != EOF) { if(m == 0 && t == 0 && 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; for(int j = 1; j <= m; j ++) { dp[i][j][0] = dp[i][j - 1][0] * (1 - p[i][j]); for(int 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]); } for(int j = 0; j <= m; j ++) s[i][j] = dp[i][m][j]; } for(int i = 1; i <= t; i ++) for(int j = 2; j <= m; j ++) s[i][j] = s[i][j] + s[i][j - 1]; double p1 = 1; for(int i = 1; i <= t; i ++) p1 *= (1 - s[i][0]); double p2 = 1; for(int i = 1; i <= t; i ++) p2 *= s[i][n - 1]; if(n == 1) p2 = 0; printf("%0.3lf ", p1 - p2); } return 0; }
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