期望dp

http://acm.hdu.edu.cn/showproblem.php?pid=4405

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6922    Accepted Submission(s): 4296

Problem Description

Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are 1,2,3,4,5,6). When Hzz is at grid i and the dice number is x, he will moves to grid i+x. Hzz finishes the game when i+x is equal to or greater than N.

There are also M flight lines on the chess map. The i-th flight line can help Hzz fly from grid Xi to Yi (0<Xi<Yi<=N) without throwing the dice. If there is another flight line from Yi, Hzz can take the flight line continuously. It is granted that there is no two or more flight lines start from the same grid.

Please help Hzz calculate the expected dice throwing times to finish the game.

 

Input

There are multiple test cases.
Each test case contains several lines.
The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000).
Then M lines follow, each line contains two integers Xi,Yi(1≤Xi<Yi≤N).  
The input end with N=0, M=0.

 

Output

For each test case in the input, you should output a line indicating the expected dice throwing times. Output should be rounded to 4 digits after decimal point.

 

Sample Input

2 0 8 3 2 4 4 5 7 8 0 0

 

Sample Output

1.1667 2.3441

 

Source

2012 ACM/ICPC Asia Regional Jinhua Online

 

Recommend

zhoujiaqi2010

题意：在一条0到n的数轴上，从0开始走到n结束，两种行走方式，第一走是摇色子决定走的步数，第二种是从一点直接飞到另一点。并给出m条飞行路线。求摇色子的期望值。

解法：从终点倒推至0，遇到飞行起点直接赋值为飞行终点的值，其他点根据期望求法一一求解。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <stdio.h>
#define INF  0x3f3f3f3f
using namespace std;
typedef long long ll ;
double dp[100009];
int vis[100009];
int main()
{
    int n , m;
    while(~scanf("%d%d" , &n , &m) && n+m)
    {
        memset(vis , 0 , sizeof(vis));
        for(int i = 1 ; i <= m ; i++)
        {
            int x , y ;
            scanf("%d%d" , &x , &y);
            vis[x] = y ;
        }
        memset(dp , 0 , sizeof(dp));
        dp[n] = 0 ;
        for(int i = n - 1 ; i >= 0 ; i--)
        {
            if(!vis[i])
            {
                for(int j = 1 ; j <= 6 ; j++)
                {
                    dp[i] += dp[i+j]/6.0;
                }
                dp[i] += 1 ;
            }
            else
                dp[i] = dp[vis[i]];
        }
        printf("%.4lf
" , dp[0]);
    }
    return 0 ;
}