# HDU 4405 Aeroplane chess （概率DP）——2012 ACM/ICPC Asia Regional Jinhua Online

2016-10-25

Aeroplane chessTime Limit: 2000 1000 MS (Java Others)Memory Limit: 32768 32768 K (Java Others)Total Submission(s): 3520Accepted Submission(s): 2232Problem Description Hzz

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3520Accepted Submission(s): 2232

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

Input There are multiple test cases.
Each test case contains several lines.
The first line contains two integers N(1&le;N&le;100000) and M(0&le;M&le;1000).
Then M lines follow, each line contains two integers Xi,Yi(1&le;XiThe 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```

MyCode：

```#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXN = 1e5+5;
int vis[MAXN];
double dp[MAXN];
int main()
{
///freopen("in.txt","r",stdin);
int n, m, x, y;
while(~scanf("%d%d",&n,&m)){
if(n==0 && m==0)
break;
memset(vis, 0, sizeof(vis));
for(int i=1; i<=m; i++){
scanf("%d%d",&x,&y);
vis[x] = y;
}
dp[n] = 0;
for(int i=n-1; i>=0; i--){
if(!vis[i]){
double sum = 0;
for(int j=1; j<=6; j++)
if(i+j < n)
sum += dp[i+j];
dp[i] = sum/6+1;
}
else
dp[i] = dp[vis[i]];
}
printf("%.4f\n",dp[0]);
}
return 0;
}```