【POJ - 2151】Check the difficulty of problems（概率dp）-CFANZ编程社区

题干：

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

题目大意：

ACM比赛中，共M道题，T个队，pij表示第i队解出第j题的概率，问每队至少解出一题且冠军队至少解出N道题的概率。

解题报告：

假设ans1为每个队至少解出一题的概率。

假设ans2为每个队解出的题数均在 [1.n-1] 之间的概率。

这样最终答案就是ans1-ans2。

ans1可以在读入的数据中预处理出来，ans2通过dp求解。

dp[i][j][k]表示第i个队在第j个题中解出k个题的概率。
求出后累加一下就好了。注意别忘dp[][j][0]需要单独处理。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int m,t,n;//m<=30,t<=1000
double p[1005][33];
double dp[1005][33][33];//dp[i][j][k]第i个队在第j个题中解出k个题的概率
double po,ans1,ans2;//ans1是所有人都做出一个题的概率 
int main()
{
  while(~scanf("%d%d%d",&m,&t,&n) && m+t+n) {
    ans1=ans2=1;
    for(int i = 1; i<=t; i++) {
      po=1;
      for(int j = 1; j<=m; j++) scanf("%lf",&p[i][j]),po *= (1-p[i][j]);
      ans1 *= (1-po);
    }
    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<=m; 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 i = 1; i<=t; i++) {
      po=0;
      for(int j = 1; j<=n-1; j++) po += dp[i][m][j];
      ans2 *= po;
    }
    printf("%.3f\n",ans1-ans2);
  }
  return 0 ;
}