poj 2151 Check the difficulty of problems（概率dp）

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6653		Accepted: 2889

Description

Organizing a programming contest(竞赛) 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(初步的) contest, the organizer can estimate(估计) the probability(可能性) that a certain team can successfully solve a certain problem.

Given the number of contest(竞赛) 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(承担) that team i solves problem j with the probability(可能性) Pij (1 <= i <= T, 1<= j <= M). Well, can you calculate(计算) 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(投入) consists of several test cases. The first line of each test case contains three integers(整数) 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(表明) the end of input, and should not be processed.

Output

For each test case, please output(输出) the answer in a separate line. The result should be rounded to three digits(数字) after the decimal(小数) point.

Sample Input

2 2 2
0.9 0.9
1 0.9
0 0 0

Sample Output

0.972

题意：
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。

解析：DP
设dp[i][j][k]表示第i个队在前j道题中解出k道的概率
则：
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]);

最后的答案就是P1-P2
*/

#include<stdio.h>
#include<string.h>
double dp[1100][32][32];
double s[1100][32];
double p[1100][32];
int M,T,N;

int main()
{
    int i,j,k;

    while(~scanf("%d%d%d",&M,&T,&N))
    {
        memset(dp,0,sizeof(dp));
        memset(s,0,sizeof(s));
        memset(p,0,sizeof(p));
        if(!M&&!T&&!N) break;
        for(i=1;i<=T;i++)
        {
            for(j=1;j<=M;j++)
                scanf("%lf",&p[i][j]);
        }

        for(i=1;i<=T;i++)//第i队 在零个题里做出0题的概率 当然为1
            dp[i][0][0]=1;

        for(i=1;i<=T;i++)//第i队 在j个题里做出0题的概率
        {
            for(j=1;j<=M;j++)
                dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]);
        }

        for(i=1;i<=T;i++)//i队 在前j个题里做出k个题的概率  其实以上都是初始化  这里才是核心
                   
        {
            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];
                }
            }

        }

        //
        for(i=1;i<=T;i++)//第i队做出0题的概率，相当于初始化
           s[i][0]=dp[i][M][0];

        for(i=1;i<=T;i++)//第i队做出1-j题的概率
          for(j=1;j<=M;j++)
            s[i][j]+=dp[i][M][j]+s[i][j-1];
            
            double p=1.0,q=1.0;
            for(i=1;i<=T;i++)
            {
                p*=(s[i][M]-s[i][0]);//每对至少做对1道题
                q*=(s[i][N-1]-s[i][0]);//每对做出1到n-1题的概率
            }

        printf("%.3f\n",p-q);

    }
    return 0;
}