POJ 2151 Check the difficulty of problems（概率dp）-优快云博客

本文链接：https://blog.youkuaiyun.com/u014737310/article/details/44344341

本文介绍了一种解决特定类型竞赛编程问题的方法，该问题涉及计算每个团队至少解决一个问题且冠军团队至少解决一定数量问题的概率。文章提供了一个示例代码，详细展示了如何通过动态规划来求解此类问题。

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Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5419		Accepted: 2384

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

0.972

Source

POJ Monthly,鲁小石

题意: n个人，m道题，求每个人做出题，并且至少一个做出k道题的概率

思路:等于每个人做出题的概率--每个人做出题并且没有一个人做出k到题的概率

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i < b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 1005

double dp[N][35][35];  // dp[i][j][k]  第 i 个人 前 j 道题 做 k 道概率
double p[N][35];
int n,m,k;

void solve()
{
	int i,j,t;
	mem(dp,0);

    fre(i,1,n+1)         //初始每一个人第一道题 对 与不对
      {
      	dp[i][1][0]=1-p[i][1];
		dp[i][1][1]=p[i][1];
      }

	fre(i,1,n+1)         //每一个人m到题都没有作对
	 {
	 	fre(j,2,m+1)
		  dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]);
	 }

	fre(i,1,n+1)        //每一个人j道题 做t道概率
	 fre(j,2,m+1)
	  fre(t,1,j+1)
	 {
       if(t<j)
		 dp[i][j][t]=dp[i][j-1][t]*(1-p[i][j])+dp[i][j-1][t-1]*p[i][j];
	   else
	     dp[i][j][t]=dp[i][j-1][t-1]*p[i][j];
	 }

	 double p1=1,p2;

	 fre(i,1,n+1)
	    p1*=(1-dp[i][m][0]);    //每个人都做出提的概率

	 p2=1;

	 fre(i,1,n+1)
	  {
	  	 double te=0;
	  	 fre(j,1,k)
	  	   te+=dp[i][m][j];
		 p2*=te;            //每个人做出题但是没有一个做了k道题的概率
	  }

	pf("%.3lf\n",p1-p2);
}

int main()
{
	int i,j;
	while(~sfff(m,n,k),n+m+k)
	{
		fre(i,1,n+1)
		   fre(j,1,m+1)
		     scanf("%lf",&p[i][j]);
	   solve();
	}
    return 0;
}