Aeroplane chess(期望dp)

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博客围绕飞机棋问题展开，棋盘有N+1个格子，从0到N编号，玩家从0格出发，掷骰子决定移动步数，棋盘还有M条飞行线可直接从Xi格飞到Yi格。通过动态规划求解走到终点的期望，给出状态转移方程。

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题目链接:https://cn.vjudge.net/problem/HDU-4405

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

Sample Output

1.1667
2.3441

题意：有n个格子，掷色子的掷出的数目就是你一次到移动格数。其中有m个飞行通道可以让你直接从第xi格

飞到第yi格。问你走到终点的期望是多少。

思路： dp[i] 表示当前在第i格走到终点的期望。

当i不是飞行通道的起点时： dp[i] =1.0+ (dp[i+1]+dp[i+2]+dp[i+3]+dp[i+4]+dp[i+5]+dp[i+6])/6.0；

当i是飞行通道的起点时： dp[Xi] = dp[Yi] (因为Xi可直接到Yi，并且不需要掷色子)；

#include<cstdio>
#include<cstring>
using namespace std;
const int N=1e5+10;
double dp[N];
int pre[N];
int main() {
	int n,m;
	while(~scanf("%d%d",&n,&m)) {
		if(n==0&&m==0) break;
		memset(pre,-1,sizeof(pre));
		memset(dp,0,sizeof(dp));
		int x,y;
		while(m--) {
			scanf("%d%d",&x,&y);
			pre[x]=y;
		}
		for(int i=n-1; i>=0; i--) {
			if(pre[i]==-1) {
				for(int j=1; j<=6; j++) {
					dp[i]+=(dp[i+j]+1)/6.0;
				}
			} else
				dp[i]=dp[pre[i]];
		}
		printf("%.4f\n",dp[0]);
	}
	return 0;
}