POJ 2151 Check the difficulty of problems （概率DP）

2019-11-06 08:13:14

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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

2 2 20.9 0.91 0.90 0 0

Sample Output

0.972

题意

ACM比赛中，共 M $M$ 道题，T $T$ 个队， p[i][j] $p[i][j]$ 表示第 i $i$ 队解出第 j $j$ 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。

思路

首先，每队至少解出一题中一定包含所有队解题数目至少为1并且都是少于 n $n$ 的，如果减去这一部分便是至少有一个队伍的解题数目大于 n $n$ 了，至于冠军队嘛！我们不用关心，因为我们只需要让它大于 n $n$ 就行了。

ans: 所有队至少解出一题的概率as : 所有队至少解出一题但是不超过 n−1 $n-1$ 题的概率

最终答案： ans−as $ans-as$

一个队伍解出不超过 n−1 $n-1$ 道题目的概率是解出 1、2、...、n−1 $1、2、...、n-1$ 道题的概率之和。

pd[j][k] $pd[j][k]$ 表示在前 j $j$ 道题中做对 k $k$ 道的概率，于是有

pd[j][k]=pd[j−1][k]∗(1.0−p[i][j])+pd[j−1][k−1]∗p[i][j] $pd[j][k]=pd[j-1][k]*(1.0-p[i][j])+pd[j-1][k-1]*p[i][j]$

其中考虑 j $j$ 题，要么做对，要么做错，只有这两种情况，其概率相加。

AC 代码

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>using namespace std;#include<queue>#include<stack>double pd[35][35],p[1005][35]; //pd[i][j]表示在前i道题中做对j道的概率int main(){ int m,t,n; while(~scanf("%d%d%d",&m,&t,&n)&&(m||t||n)) { double ans=1.0,as=1.0; //ans 为所有队至少A一道题的概率 as为所有队A的题目数量都不超过n的概率 memset(pd,0,sizeof(pd)); for(int i=0; i<t; i++) { double temp=1.0; for(int j=1; j<=m; j++) { scanf("%lf",&p[i][j]); temp*=1.0-p[i][j]; //逆命题就是该队伍一道题也没有做对 } ans*=1.0-temp; //因为要保证所有队伍都满足，所以用乘法定理 } for(int i=0; i<t; i++) { memset(pd,0,sizeof(pd)); pd[0][0]=1.0; //在前0道题目中A0道永真 for(int j=1; j<=m; j++) { pd[j][0]=pd[j-1][0]*(1-p[i][j]); for(int k=1; k<=j; k++) pd[j][k]=pd[j-1][k]*(1.0-p[i][j])+pd[j-1][k-1]*p[i][j]; //第j题要么做对，要么做错 } double temp=0.0; for(int j=1; j<n; j++) temp+=pd[m][j]; //前m道题中做对数目少于n的子命题（做对1、2...n-1道）之间用加法定理 as*=temp; //各组之间乘法定理 } printf("%.3f/n",ans-as); } return 0;}

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