hdu4405概率dp入门

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1122    Accepted Submission(s): 762

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

 

/*题意:有一个飞行棋n个格子，刚開始在0这个位置，每次能够扔色子,扔到x则能够移动x格
假设到达的位置>=n则胜利
另外在棋盘上还有m对点x,y表示在位置x能够直接飞行到y
求到达胜利平均的扔色子次数

分析:求期望,假设dp[i]表示在点i位置到达胜利所须要的平均次数,则我们须要求dp[0]
dp[n],dp[n+1]....等是知道的:为0
而对于dp[i]能够到达:
假设点i不能飞行：dp[i+1],dp[i+2],dp[i+3],dp[i+4],dp[i+5],dp[i+6]且等概率:1/6
假设i点位置能够飞行则能够到达飞行的点
由E(aA+bB+cC+dD...)=aEA+bEB+....
可知:dp[i]=1/6*dp[i+1]+1/6*dp[i+2]...
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=100000+10;
int n,m,size;
int head[MAX];
double dp[MAX];//dp[i]表示从i到达目标所须要的平均次数(期望)

struct Node{
	int y,next;
	Node(){}
	Node(int Y,int NEXT):y(Y),next(NEXT){}
}node[MAX];

void Init(){
	memset(head,-1,sizeof head);
	memset(dp,0,sizeof dp);
	size=0;
}

void InsertNode(int x,int y){
	node[size]=Node(y,head[x]);
	head[x]=size++;
}

double Solve(int x){
	double sum=0,num=0;
	for(int i=head[x];i != -1;i=node[i].next){
		sum+=dp[node[i].y];
		++num;
	}
	return sum/num;
}

int main(){
	int x,y;
	while(~scanf("%d%d",&n,&m),n+m){
		Init();
		for(int i=0;i<m;++i){
			scanf("%d%d",&x,&y);
			InsertNode(x,y);
		}
		for(int i=n-1;i>=0;--i){
			if(head[i] != -1){
				dp[i]=Solve(i);
			}else{
				dp[i]=(dp[i+1]+dp[i+2]+dp[i+3]+dp[i+4]+dp[i+5]+dp[i+6])/6+1;
			}
		}
		printf("%.4lf\n",dp[0]);
	}
	return 0;
}