HDU 4405 期望dp

Problem 

DescriptionHzz 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.InputThere 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.OutputFor 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

分析：真没看懂他那个飞来飞去的航线是什么鬼，必须飞？可以飞？，好吧姑且算他必须飞吧，否则对不起出题人。从后往前，不难看出f[i]表示在i位置离目标的期望，因为扔了骰子过后他可以到后面1~6的点，每点的概率为1/6，故而加上就ok了。若他可以飞到某点，则直接f[i]=f[to[i]]，必须飞。所以转移就出来了。

# include <iostream># include <cstdio># include <cmath># include <list># include <cstring># include <map># include <ctime># include <algorithm># include <queue>using namespace std;typedef long long ll;int read(){int f=1,i=0;char ch=getchar();while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9') {i=(i<<3)+(i<<1)+ch-'0';ch=getchar();}return f*i;}int n,m,a,b;double f[100005];int to[100005];int main(){    int n,m;    while(scanf("%d%d",&n,&m)&&(n||m)){        memset(to,-1,sizeof(to));        for(int i=1;i<=m;i++){            a=read(),b=read();            to[a]=b;        }        memset(f,0,sizeof(f));        for(int i=n-1;i>=0;i--){            if(to[i]==-1){                for(int j=1;j<=6;j++)    f[i]+=f[i+j]/6.0;                f[i]+=1;            }else f[i]=f[to[i]];        }        printf("%0.4f\n",f[0]);    }    return 0;}