poj 2151 Check the difficulty of problems(概率dp)
2015-04-23 09:19
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题目链接
Check the difficulty of problems
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
Sample Output
题意:M道题目,T个人,已经每个人做出每道题目的概率,求所有人都过题,但是最多过N到题的概率。
题解:首先用dp[i][j][k]表示第i个人对于前j个题目,做出来k道题目的概率。通过这个dp我们先求出第i个人过j道题目的概率。然后再用f[i][j]表示前i个人,都过了题,最多过了j道题的概率。最后统计答案即可。
代码如下:
Check the difficulty of problems
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 5482 | Accepted: 2414 |
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
题意:M道题目,T个人,已经每个人做出每道题目的概率,求所有人都过题,但是最多过N到题的概率。
题解:首先用dp[i][j][k]表示第i个人对于前j个题目,做出来k道题目的概率。通过这个dp我们先求出第i个人过j道题目的概率。然后再用f[i][j]表示前i个人,都过了题,最多过了j道题的概率。最后统计答案即可。
代码如下:
#include<stdio.h> #include<iostream> #include<algorithm> #include<set> #include<vector> #include<string.h> #include<map> #define inff 0x3fffffff #define nn 1100 #define mod 1000003 typedef __int64 LL; typedef unsigned __int64 LLU; using namespace std; int m,t,n; double p[nn][33]; double dp[nn][33][33]; double sum[nn][33]; double f[nn][33]; int main() { int i,j,k; while(scanf("%d%d%d",&m,&t,&n)&&(m+t+n)) { for(i=1;i<=t;i++) { for(j=1;j<=m;j++) { scanf("%lf",&p[i][j]); } } memset(dp,0,sizeof(dp)); for(i=1;i<=t;i++) { dp[i][0][0]=1.0; for(j=1;j<=m;j++) { for(k=0;k<=j;k++) { dp[i][j][k]+=dp[i][j-1][k]*(1-p[i][j]); if(k>0) dp[i][j][k]+=dp[i][j-1][k-1]*p[i][j]; } } } for(i=1;i<=t;i++) { sum[i][0]=dp[i][m][0]; for(j=1;j<=m;j++) { sum[i][j]=sum[i][j-1]+dp[i][m][j]; } } memset(f,0,sizeof(f)); f[0][0]=1.0; for(i=1;i<=t;i++) { for(j=1;j<=m;j++) { f[i][j]+=f[i-1][j]*(sum[i][j]-sum[i][0]); for(k=0;k<j;k++) { f[i][j]+=f[i-1][k]*(sum[i][j]-sum[i][j-1]); } } } double ans=0; for(i=n;i<=m;i++) { ans+=f[t][i]; } printf("%.3lf\n",ans); } return 0; }
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