POJ 2151 Check the difficulty of problems
2015-05-08 23:50
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题目链接:POJ 2151 Check the difficulty of problems
题面:
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
Source
POJ Monthly,鲁小石
第一次写概率DP,感觉和DP有些不一样,借鉴了这篇博客:金海峰
代码:
题面:
Check the difficulty of problems
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 5519 | Accepted: 2431 |
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
Source
POJ Monthly,鲁小石
第一次写概率DP,感觉和DP有些不一样,借鉴了这篇博客:金海峰
代码:
#include <iostream> #include <cstdio> #include <cstring> using namespace std; double correspond[1005][31]; double dp[1005][31][31]; int main() { int n,m,t; double ans=1,tmp,sum; while(scanf("%d%d%d",&m,&t,&n)&&(m||t||n)) { ans=tmp=1; for(int i=0;i<t;i++) { for(int j=1;j<=m;j++) scanf("%lf",&correspond[i][j]); } memset(dp,0,sizeof(dp)); //dp[i][j][k] 第i只队伍,做前j题,做对k题的概率 for(int i=0;i<t;i++) { dp[i][0][0]=1; for (int j=1;j<=m;j++) { //还是做对0道,那么就是当前这一道没做对 dp[i][j][0] = dp[i][j -1][0] * (1- correspond[i][j]); for (int k =1; k <= j; k++) //做对k道,要么是之前就做对了k道,这次没对,要么就是之前对了k-1道,这次对了 dp[i][j][k] = dp[i][j -1][k -1] * correspond[i][j]+ dp[i][j -1][k] * (1- correspond[i][j]); } } //排除任何一个人一道都没做对的情况 for(int i=0;i<t;i++) ans *=(1- dp[i][m][0]); //排除所有人都只做对1~n-1的情况 for(int i=0;i<t;i++) { sum=0; for(int j=1;j<n;j++) sum+=dp[i][m][j]; tmp*=sum; } ans-=tmp; printf("%.3f\n", ans); } return 0; }
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