[ACM] POJ 2151 Check the difficulty of problems (概率+DP)
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Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 4748 | Accepted: 2078 |
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
这道题太无语了...特别不好理解...... 概率,概率,概率......
题意为 有T个队參加ACM比赛。一共同拥有M道题,问全部的队都至少做出一道题且冠军队做出的题目不能少于N道的概率。
dp[1010][32][32];//dp[i][j][k]表示第i个队前j道题中做出k道
p[1010][32];//p[i][j] 表示第i个队做出第j道题的概率
s[1010][32];//s[i][j]表示第i各队做出的题目小于等于j道的概率
转载于:http://www.cnblogs.com/kuangbin/archive/2012/10/02/2710606.html
设dp[i][j][k]表示第i个队在前j道题中解出k道的概率
则:
dp[i][j][k]=dp[i][j-1][k-1]*p[j][k]+dp[i][j-1][k]*(1-p[j][k]);
先初始化算出dp[i][0][0]和dp[i][j][0];
设s[i][k]表示第i队做出的题小于等于k的概率
则s[i][k]=dp[i][M][0]+dp[i][M][1]+``````+dp[i][M][k];
则每一个队至少做出一道题概率为P1=(1-s[1][0])*(1-s[2][0])*```(1-s[T][0]);
每一个队做出的题数都在1~N-1的概率为P2=(s[1][N-1]-s[1][0])*(s[2][N-1]-s[2][0])*```(s[T][N-1]-s[T][0]);
最后的答案就是P1-P2
代码:
#include <iostream> #include <stdio.h> #include <string.h> using namespace std; double dp[1010][32][32];//dp[i][j][k]表示第i个队前j道题中做出k道 double p[1010][32];//p[i][j] 表示第i个队做出第j道题的概率 double s[1010][32];//s[i][j]表示第i各队做出的题目小于等于j道的概率 int m,n,t; int main() { while(scanf("%d%d%d",&m,&t,&n)!=EOF&&(m||t||n)) { 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]);//第i个队前j道题目都做不出来的概率 for(int j=1;j<=m;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]); //第i个队前j道题做出k道的概率等于前j-1道题做出k-1道,第j道做出来加上前j-1道做出k道,第j道没做出来的概率和 s[i][0]=dp[i][m][0]; for(int j=1;j<=m;j++) s[i][j]=s[i][j-1]+dp[i][m][j];//依据上面的s数组求法可得递推公式 } double p1=1,p2=1; for(int i=1;i<=t;i++) { p1*=(1-s[i][0]);//全部的队至少做出一题的概率 p2*=(s[i][n-1]-s[i][0]);//全部的队都做出来1到n-1道题 } printf("%.3lf\\n",p1-p2); } return 0; }
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