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  • E. Fish (概率DP)

    E. Fish
    time limit per test
    3 seconds
    memory limit per test
    128 megabytes
    input
    standard input
    output
    standard output

    n fish, numbered from 1 to n, live in a lake. Every day right one pair of fish meet, and the probability of each other pair meeting is the same. If two fish with indexes i and j meet, the first will eat up the second with the probability aij, and the second will eat up the first with the probability aji = 1 - aij. The described process goes on until there are at least two fish in the lake. For each fish find out the probability that it will survive to be the last in the lake.

    Input

    The first line contains integer n (1 ≤ n ≤ 18) — the amount of fish in the lake. Then there follow n lines with n real numbers each — matrix aaij (0 ≤ aij ≤ 1) — the probability that fish with index i eats up fish with index j. It's guaranteed that the main diagonal contains zeros only, and for other elements the following is true: aij = 1 - aji. All real numbers are given with not more than 6 characters after the decimal point.

    Output

    Output n space-separated real numbers accurate to not less than 6 decimal places. Number with index i should be equal to the probability that fish with index i will survive to be the last in the lake.

    Sample test(s)
    input
    2
    0 0.5
    0.5 0
    output
    0.500000 0.500000 
    input
    5
    0 1 1 1 1
    0 0 0.5 0.5 0.5
    0 0.5 0 0.5 0.5
    0 0.5 0.5 0 0.5
    0 0.5 0.5 0.5 0
    output
    1.000000 0.000000 0.000000 0.000000 0.000000 

     1 #include <iostream>
     2 #include <string.h>
     3 #include <stdio.h>
     4 using namespace std;
     5 double a[18][18],b[1<<18];
     6 int fun(int x)
     7 {
     8     int s=0;
     9     while(x)
    10     {
    11         s+=x&1;
    12         x>>=1;
    13     }
    14     return s;
    15 }
    16 int main()
    17 {
    18     int n,i,r,t,j;
    19     cin>>n;
    20     for(i=0; i<n; i++)for(j=0; j<n; j++)scanf("%lf",&a[i][j]);
    21     memset(b,0,sizeof(b));
    22     b[(1<<n)-1]=1;
    23     for(i=(1<<n)-1; i>=0; i--)
    24     {
    25         int c=fun(i);
    26         c=c*(c-1)/2;
    27         for(r=0;r<n;r++)
    28         if(i&(1<<r))
    29         for(t=0;t<n;t++)
    30         if(i&(1<<t))
    31         b[i-(1<<t)]+=b[i]*a[r][t]/c;
    32     }
    33     for(r=0;r<n-1;r++)
    34     printf("%.6lf ",b[1<<r]);printf("%.6lf
    ",b[1<<r]);
    35 }
    View Code
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  • 原文地址:https://www.cnblogs.com/ERKE/p/3586220.html
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