hdu 4405　全期望公式

首先确定随机事件Ｘ：掷色子到达终点的次数

Ｅ［n-n+5]　：因为已经到达终点所以，全为０

由全期望公式得，E[i] = sum( E[i+j] ) ( 1<=j<=6 )/6 + 1因为是等概率事件

如果有飞机的话，就是E[i] = E[to[i]]

递推求取即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX 100017

using namespace std;

double dp[MAX];
int to[MAX];

int n , m , a, b;

int main  ( )
{
    while ( ~scanf ( "%d%d" , &n , &m ) , n+m )
    {
        for ( int i = 0 ; i < MAX ; i++ )
            dp[i] = to[i] = 0;
        while ( m-- )
        {
            scanf ( "%d%d" , &a , &b );
            to[a] = b;   
        }
        for ( int i = n-1 ; i >= 0 ; i-- )
        {
            if ( to[i] ) dp[i] = dp[to[i]];
            else
            {
                for ( int j = 1 ; j <= 6 ; j++ ) dp[i] += dp[i+j];
                dp[i] /= 6.0;
                dp[i] += 1.0;
            }       
        }
        printf ( "%.4lf\n" , dp[0] );
    }
}