Check the difficulty of problems（POJ 2151）

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5457		Accepted: 2400

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

概率dp

开始理解错了题意，弄了半天也没搞出来，后来理解完了啦，发现概率都忘了。

然后补概率。

设A = “所有队都至少做完一题”， B = “至少存在一个队做完不少于n题”；

P(A) = P(A(B + !B)) = P(AB) + P(A!B);

P(AB) = P(A) - P(A!B);

#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define ll long long
#define _cle(m, a) memset(m, a, sizeof(m))
#define repu(i, a, b) for(int i = a; i < b; i++)
#define MAXN 1005
double p[MAXN][35];
double d[MAXN][35][35];
double s[MAXN][35];
int main()
{
    int m, t, n;
    while(~scanf("%d%d%d", &m, &t, &n) && (m + t + n))
    {
        for(int i = 0; i < t; i++)
            for(int j = 1; j <= m; j++)
            scanf("%lf", &p[i][j]);
        _cle(d, 0);

        for(int i = 0; i < t; i++) {
            d[i][0][0] = 1.0;
            for(int j = 1; j <= m; j++)
                d[i][j][0] = d[i][j - 1][0] * (1.0 - p[i][j]);
        }

        for(int i = 0; i < t; i++)
            for(int j = 1; j <= m; j++)
               for(int k = 1; k <= j; k++)
                 d[i][j][k] += (d[i][j - 1][k - 1] * p[i][j] + d[i][j - 1][k] * (1.0 - p[i][j]));

        double p1 = 1.0, p2 = 1.0;

        for(int i = 0; i < t; i++) {
            s[i][0] = d[i][m][0];
            for(int j = 1; j <= m; j++) s[i][j] = s[i][j - 1] + d[i][m][j];
        }

        for(int i = 0; i < t; i++) p1 *= (s[i][m] - s[i][0]);
        for(int i = 0; i < t; i++) p2 *= (s[i][n - 1] - s[i][0]);
        printf("%.3lf
", p1 - p2);
    }
    return 0;
}

View Code