hdu 4405 Aeroplane chess （概率DP）

Aeroplane chess

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

Total Submission(s): 1503 Accepted Submission(s): 1025



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

学习概率DP推荐一个链接：http://kicd.blog.163.com/blog/static/126961911200910168335852/

思路：由当前点能够走向以下6个相邻位置，走到这几个点的概率均相等。用dp[i]表示该点走到目标的期望步数，则该点的期望能够由它能够到达的6个点相加得到，由于它走到下一个位置花费时间1，故要加一。见式子：

dp[0]=dp[1]*1/6+dp[2]*1/6+dp[3]*1/6+dp[4]*1/6+dp[5]*1/6+dp[6]*1/6+1;
dp
=0(自身到自身期望为0)

那么，我们倒着推过来就能得到答案为dp[0]。

#include"stdio.h"
#include"string.h"
#include"iostream"
#include"algorithm"
#include"math.h"
#include"vector"
using namespace std;
#define LL __int64
#define N 100005
#define max(a,b) (a>b?

a:b)
vector<int>g
;
int vis
;
double dp
;
int main()
{
int n,m,i,j,v,a,b;
while(scanf("%d%d",&n,&m),n||m)
{
for(i=0;i<=n;i++)
g[i].clear();
for(i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
g[b].push_back(a);
}
memset(dp,0,sizeof(dp));      //易知dp
=0
memset(vis,0,sizeof(vis));
for(i=0;i<g
.size();i++)
{
v=g
[i];
dp[v]=dp
;
vis[v]=1;
}
for(i=n-1;i>=0;i--)
{
if(!vis[i])
{
for(j=i+1;j<=i+6;j++)
{
dp[i]+=dp[j]/6;
}
dp[i]+=1;
}
for(j=0;j<g[i].size();j++)
{
v=g[i][j];
dp[v]=dp[i];
vis[v]=1;
}
}
printf("%.4f\n",dp[0]);
}
return 0;
}