Check the difficulty of problems
| Time Limit: 2000MS |
|
Memory Limit: 65536K |
| Total Submissions: 8903 |
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Accepted: 3772 |
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 2
0.9 0.9
1 0.9
0 0 0
Sample Output
0.972
Source
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\n", p1 - p2);
}
return 0;
}
【POJ】2151：Check the difficulty of problems【概率DP】
原文：https://www.cnblogs.com/wans-caesar-02111007/p/9835722.html
