hdu 4405 Aeroplane chess 概率dp

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1499    Accepted Submission(s): 1022

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   |   We have carefully selected several similar problems for you:  5041 5040 5039 5038 5037 

 

題意：飛行棋，擲骰子決定走的點數，其中還有一些近路，不用花費點數就能通過，求走到n的期望。

思路：把邊記錄下來，邊兩邊的期望值相等，從後向前掃一遍。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
using namespace std;
const int mod=1e9+7;
const int N=100010;
double dp[N];
int edge[N];
int main(){
    int n,m,a,b;
    while(scanf("%d%d",&n,&m),n||m){
        memset(dp,0,sizeof(dp));
        memset(edge,0,sizeof(edge));
        while(m--){
            scanf("%d%d",&a,&b);
            edge[a]=b;
        }
        double p=1.0/6;
        for(int i=n-1;i>=0;i--){
            if(edge[i]){
                dp[i]=dp[edge[i]];
                continue;
            }
            dp[i]=1;
            for(int j=1;j<=6;j++){
                dp[i]+=dp[i+j]*p;
            }
        }

    printf("%.4lf\n",dp[0]);
    }
    return 0;
}