• POJ 2151 Check the difficulty of problems


    Check the difficulty of problems
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 4011   Accepted: 1779

    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

    Source

    POJ Monthly,鲁小石 

    设a[i][j][k]表示第i队在前j道题中共解出k道题的概率,易得a[i][j][k]有如下递推
    关系(另需考虑边界条件):
    a[i][j][k] = a[i][j-1][k-1] * p[i][j] + a[i][j-1][k] * (1-p[i][j])
    设s[i][j]表示a[i][M][0] + a[i][M][1] + ... + a[i][M][j]
    问题的解可以转化为:每队均至少做一题的概率(用P1表示)减去每队做题数均在1到N-1
    之间的概率(用P2表示)。
    P1 = (s[1][M] - s[1][0])*(s[2][M]-s[2][0])*...*(s[T][M]-s[T][0])
    P2 = (s[1][N-1] - s[1][0])*(s[2][N-1]-s[2][0])*...*(s[T][N-1]-s[T][0])

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 
     5 using namespace std;
     6 
     7 double dp[1100][40][40];
     8 double p[1100][40];
     9 double sum[1100][40];
    10 double p1,pn;
    11 int M,T,N;
    12 
    13 int main()
    14 {
    15     while(scanf("%d%d%d",&M,&T,&N)!=EOF)
    16     {
    17         if((M||T||N)==0) break;
    18 
    19         memset(dp,0,sizeof(dp));
    20         memset(sum,0,sizeof(sum));
    21         memset(p,0,sizeof(p));
    22 
    23         for(int i=1;i<=T;i++)
    24         {
    25             p[i][0]=1;
    26             for(int j=1;j<=M;j++)
    27             {
    28                 scanf("%lf",&p[i][j]);
    29             }
    30         }
    31 
    32         for(int i=1;i<=T;i++)
    33         {
    34             dp[i][0][0]=1;
    35         }
    36 
    37         for(int i=1;i<=T;i++)
    38         {
    39             for(int j=1;j<=M;j++)
    40             {
    41                 for(int k=0;k<=j;k++)
    42                 {
    43                     dp[i][j][k]=dp[i][j-1][k-1]*p[i][j]+dp[i][j-1][k]*(1-p[i][j]);
    44                 }
    45             }
    46         }
    47 
    48         for(int i=1;i<=T;i++)
    49         {
    50             for(int j=0;j<=M;j++)
    51             {
    52                 for(int k=0;k<=j;k++)
    53                     sum[i][j]+=dp[i][M][k];
    54             }
    55         }
    56 /*
    57         for(int i=1;i<=T;putchar(10),i++)
    58         {
    59             for(int j=1;j<=M;putchar(10),j++)
    60             {
    61                 for(int k=0;k<=M;k++)
    62                 {
    63                     printf("dp[%d][%d][%d]:   %lf
    ",i,j,k,dp[i][j][k]);
    64                 }
    65             }
    66         }
    67 
    68         for(int i=1;i<=T;putchar(10),i++)
    69         {
    70             for(int k=0;k<=M;k++)
    71             {
    72                 printf("sum[%d][%d]:  %lf
    ",i,k,sum[i][k]);
    73             }
    74         }
    75 */
    76         p1=1;pn=1;
    77         for(int i=1;i<=T;i++)
    78         {
    79             p1*=(sum[i][M]-sum[i][0]);
    80             pn*=(sum[i][N-1]-sum[i][0]);
    81         }
    82         printf("%.3lf
    ",p1-pn);
    83     }
    84 
    85     return 0;
    86 }
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  • 原文地址:https://www.cnblogs.com/CKboss/p/3314048.html
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