HDU 4405 Aeroplane chess（概率DP求期望）

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 361    Accepted Submission(s): 255

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

 

 

网赛时很简单的概率dp的题目。那个时候不会概率dp，没有做过概率dp太弱了~~~~~~~

/*

概率DP求期望。
形成一个有向无环图。按照公式递推就可以了。
dp[i]表示i点跳到目标状态的期望步数

*/


#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
using namespace std;

const int MAXN=100010;
double dp[MAXN];
vector<int>vec[MAXN];
bool used[MAXN];
int main()
{
    int n,m;
    int u,v;
    while(scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)break;
        for(int i=0;i<=n;i++)vec[i].clear();
        memset(dp,0,sizeof(dp));
        while(m--)
        {
            scanf("%d%d",&u,&v);
            vec[v].push_back(u);
        }
        memset(used,false,sizeof(used));
        for(int i=0;i<vec[n].size();i++)
        {
            v=vec[n][i];
            dp[v]=0;
            used[v]=true;
        }
        for(int i=n-1;i>=0;i--)
        {
            if(used[i]==false)
            {
                for(int j=i+1;j<=i+6;j++)dp[i]+=dp[j]/6;
                dp[i]+=1;
                used[i]=true;
            }

            for(int j=0;j<vec[i].size();j++)
            {
                v=vec[i][j];
                dp[v]=dp[i];
                used[v]=true;
            }
        }
        printf("%.4lf\n",dp[0]);
    }
    return 0;
}