poj 2151 概率DP（水）

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5750		Accepted: 2510

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

题意：在acm比赛中，n题，t队。给出每个队做对每题的概率，问每队至少对一题，至少有一队做对至少m题的概率。（本解题报告中的n，m与原题中相反）

分析：dp，f[i][j]表示第i个队伍做对第j题的概率。g[i][j][k]表示第i个队伍对于前j题而言做对k道的概率。

g[i][j][k] = g[i][j - 1][k - 1] * (f[i][j]) + g[i][j - 1][k] * (1 - f[i][j]);

有了所有的g，我们就可以求出每个队至少做对1题的概率：ans *= 1 - g[i][n][0];

再减去每个队都只做对1～m-1题的概率(把每个队做对1～m-1题的概率加和，并把各队结果相乘）

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
double f[][];
double dp[][][];
int main(){
      int n,t,m;
      while(scanf("%d%d%d",&n,&t,&m)!=EOF){
            if(n==&&t==&&m==)
            break;
         memset(f,,sizeof(f));
         memset(dp,,sizeof(dp));
         for(int i=;i<t;i++){
            for(int j=;j<=n;j++)
                scanf("%lf",&f[i][j]);
         }

         for(int i=;i<t;i++){
            dp[i][][]=;
            for(int j=;j<=n;j++){
                dp[i][j][]=dp[i][j-][]*(-f[i][j]);
                for(int k=;k<=j;k++)
                    dp[i][j][k]=dp[i][j-][k-]*f[i][j]+dp[i][j-][k]*(-f[i][j]);
            }
         }

         double ans=;
         for(int i=;i<t;i++)
            ans*=(-dp[i][n][]);

         double temp=;
         for(int i=;i<t;i++){
            double sum=;
            for(int j=;j<m;j++)
                sum+=dp[i][n][j];
            temp*=sum;
         }

         printf("%.3lf\n",ans-temp);

      }
      return ;
}