HDU 4405：Aeroplane chess（概率DP）

Aeroplane chess

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

题意：类似飞行棋，有0到N格，从0开始走，然后每掷一次骰子，就走骰子上的点数，然后棋盘上有M条捷径，当到达捷径的其实点x 时可以直接飞到捷径的终点y.问到达终点所用次数的期望，其中刚好到达N，或者超过N都算到达。比如N为2，你掷的骰子为3 ，则到达。

分析：题目叫我们求期望，应该从后往前推，定义一个数组f[i]表示第i格到达目标所需要次数的期望，然后用不定长数组vector<int>v[N+1],存储捷径的起始点，因为起始点的期望跟终点的期望相等。在每次求f[i]时，应该把f[i]赋值给能到达i的起始点

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<queue> 
#include<algorithm>
using namespace std;
int n,m;

vector <int> v[100010];

//int x[1005];
//int y[1005];
double f[100010];
int flag[100010]; 
int main ()
{
	while (scanf ("%d%d",&n,&m)!=EOF){
		if (n==0&&m==0)break;
		for (int i=0;i<=n;i++) v[i].clear();
		for (int i=1;i<=m;i++){
			int x,y;
			scanf ("%d%d",&x ,&y ); 
			v[y].push_back(x);// 把能到达y 的起始点x放在对应的终点y 对应的数组中 
		}
		memset(f,0,sizeof (f));
		memset(flag,0,sizeof (flag));
		f[n]=0;//flag[n]=1;
		for (int i=0;i<v[n].size();i++){
			int t=v[n][i];
			f[t]=f[n];//终点为n的捷径的起点的期望与终点一样
			flag[t]=1;//标记； 
		}
		for(int i=n-1;i>=0;i--){
			if (flag[i]==0){
				f[i]=(f[i+1]+f[i+2]+f[i+3]+f[i+4]+f[i+5]+f[i+6])/6+1;
				flag[i]=1;
			}
			
			for (int j=0;j<v[i].size();j++){
				int t=v[i][j];
				f[t]=f[i];
				flag[t]=1;//标记 
			}
		} 
		printf ("%.4f\n",f[0]);
	}	
	return 0;
}