POJ 2151 Check the difficulty of problems dp

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5488		Accepted: 2419

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

dp[i][j][k]表示第i个队在前j道题中解出k道的概率
dp[i][j][k]=dp[i][j-1][k-1]*p[j][k]+dp[i][j-1][k]*(1-p[j][k]);
先初始化算出dp[i][0][0]和dp[i][j][0];
设s[i][k]表示第i队做出的题小于等于k的概率
则s[i][k]=dp[i][M][0]+dp[i][M][1]+``````+dp[i][M][k];
则每个队至少做出一道题概率为P1=(1-s[1][0])*(1-s[2][0])*```(1-s[T][0]);
每个队做出的题数都在1~N-1的概率为P2=(s[1][N-1]-s[1][0])*(s[2][N-1]-s[2][0])*```(s[T][N-1]-s[T][0]);
答案P1-P2

/** Author: ¡î¡¤aosaki(*¡¯(OO)¡¯*)  niconiconi¡ï **/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <list>
#include <stack>
//#include <tuple>
#define mem(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define lp(k,a) for(int k=1;k<=a;k++)
#define lp0(k,a) for(int k=0;k<a;k++)
#define lpn(k,n,a) for(int k=n;k<=a;k++)
#define lpd(k,n,a) for(int k=n;k>=a;k--)
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define lowbit(x) (x&(-x))
#define ll long long
#define pi pair<int,int>
#define vi vector<int>
#define PI acos(-1.0)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define TT cout<<"*****"<<endl;
#define TTT cout<<"********"<<endl;
inline int gcd(int a,int b)
{
    return a==0?b:gcd(b%a,a);
}

#define INF 1e9
#define eps 1e-8
#define mod 10007
#define MAX 10010

using namespace std;

double dp[1010][35][35],p[1010][35];
double s[1010][35],x1,x2;
int m,t,n;

int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d%d%d",&m,&t,&n) && m!=0)
    {
        mem(dp);
        mem(s);
        lp(i,t)
          {
             lp(j,m)
               scanf("%lf",&p[i][j]);
          }
        lp(i,t)
        {
            dp[i][0][0]=1.0;
            lp(j,m)
          {
              dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]);
          }
        }

        lp(i,t)
         lp(j,m)
         {
             lp(k,j)
              dp[i][j][k]=dp[i][j-1][k-1]*p[i][j]+dp[i][j-1][k]*(1-p[i][j]);
         }
        lp(i,t)
        {
           s[i][0]=dp[i][m][0];
             lp(j,n)
              s[i][j]=s[i][j-1]+dp[i][m][j];
        }

        x1=1.0; x2=1.0;
        lp(i,t)
        {
           x1*=(1-s[i][0]);
           x2*=(s[i][n-1]-s[i][0]);
        }

        printf("%.3f\n",x1-x2);

    }


    return 0;
}