hdu4405(期望dp)

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4732    Accepted Submission(s): 2971

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 08 32 44 57 80 0

 

Sample Output

1.16672.3441

题目大意：在一条有n个格点的线上掷骰子，有1-6六个值，如果当前为止超过n，则胜利，有m个跳跃点，到x位置可以直接飞到y位置(x < y)，问从起点开始到终点掷骰子的期望数。

思路：从n到0反过来求，首先对于i>=n，dp[i]=0，然后对于0<=i<n，dp[i]=dp[i+1]+...+dp[i+6]+1，因为投掷一次几率是平等的。最后计算到dp[0]即可。

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;#define maxn 100005int n,m;double dp[maxn];int connect[maxn];int main(){while(~scanf("%d%d",&n,&m)&&(n||m)){memset(dp,0,sizeof(dp));memset(connect,-1,sizeof(connect));int x,y;for(int i=0;i<m;i++){scanf("%d%d",&x,&y);connect[x]=y;}for(int i=n-1;i>=0;i--){if(connect[i]!=-1)dp[i]=dp[connect[i]];else{for(int j=1;j<=6;j++)dp[i]+=(dp[i+j]/6);dp[i]++;}}printf("%.4lf\n",dp[0]);}}