POJ 2151 - Check the difficulty of problems

算法

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发布日期: 2011-08-27

更新日期: 2026-11-06

文章字数: 995

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POJ 2151 - Check the difficulty of problems
Time: 2000MS
Memory: 65536K
难度: 初级
分类: 概率

0x00 POJ 原题

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:

All of the teams solve at least one problem.
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

0x10 题目概述

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。

0x20 解题思路

概率+DP，概率不好真的拿不下这题

这题难点不在编程，在于问题的转化和理解

只要能用笔算出答案，离 AC 也就不远了。。。

要求：

每队至少解出一题且冠军队至少解出 N 道题的概率
由于冠军队可以不止一队，即允许存在并列冠军

则原来的所求的概率可以转化为：

每队均至少做一题的概率 P1 减去每队做题数均在 1 到 N-1 之间的概率 P2

0x30 AC 源码

//Memory Time
//8272K  110MS 

#include<iostream>
#include<iomanip>
using namespace std;

int M,T,N;   //M:题数  T:队数   N:冠军队至少做题数
double dp[1001][31][31];   //状态方程: dp[i][j][k]为第i队PASS前j题中的k题的概率

double p[1001][31];  //p[i][j]为第i队通过第j题的概率
double s[1001][31];  //s[i][j]为第i队在M题中至少PASS j题的概率

void ProTable(void)  //概率打表
{
    memset(dp,0.0,sizeof(dp));
    memset(s,0.0,sizeof(s));

    int i,j,k;
    for(i=1;i<=T;i++)  //逐队枚举
    {
        /*Initial*/

        dp[i][0][0]=1.0;

        for(j=1;j<=M;j++)
            dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]);

        /*Dp*/

        for(j=1;j<=M;j++)
            for(k=1;k<=j;k++)
                dp[i][j][k] = dp[i][j-1][k-1]*p[i][j] + dp[i][j-1][k]*(1-p[i][j]);

        s[i][0]=dp[i][M][0];
        for(k=1;k<=M;k++)
            s[i][k]=s[i][k-1]+dp[i][M][k];
    }
    return;
}

int main(int i,int j)
{
    while(cin>>M>>T>>N)
    {
        if(!M && !T && !N)
            break;

        /*Input*/

        for(i=1;i<=T;i++)
            for(j=1;j<=M;j++)
                cin>>p[i][j];

        /*Compute the Probability*/

        ProTable();

        double p1=1.0;
        for(i=1;i<=T;i++)
            p1*=(s[i][M]-s[i][0]);   //所有队至少做1题的概率

        double p2=1.0;
        for(i=1;i<=T;i++)
            p2*=(s[i][N-1]-s[i][0]); //所有队做的题数均在1~N-1之间的概率

        /*Output*/

        cout<<fixed<<setprecision(3)<<p1-p2<<endl;
        //每队至少解出一题 且 至少有一队(冠军队)能至少解出N道题的概率
    }
    return 0;
}