POJ 2151 Check the difficulty of problems（概率dp）

Language: Default

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

2 2 2
0.9 0.9
1 0.9
0 0 0

Sample Output

0.972

Source

POJ Monthly,鲁小石

题意: n个人，m道题，求每一个人做出题，而且至少一个做出k道题的概率

思路:等于每一个人做出题的概率--每一个人做出题而且没有一个人做出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
")

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)         //初始每个人第一道题 对 与不正确
      {
      	dp[i][1][0]=1-p[i][1];
		dp[i][1][1]=p[i][1];
      }

	fre(i,1,n+1)         //每个人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;            //每个人做出题可是没有一个做了k道题的概率
	  }

	pf("%.3lf
",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;
}