題目描述：
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
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
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
很簡單的概率dp題，幾行代碼搞定，也很容易理解，貼上代碼，算是我的概率DP入門題：
/*******************************************************************************/
/* OS : Linux fc20.x86_64 #1 SMP Tue Dec UTC 2013 x86_64 GNU/Linux
* Compiler : 4.8.2 20131212 (Red Hat 4.8.2-7) (GCC)
* Encoding : UTF8
* Date : 2014-04-02
* All Rights Reserved by alop.
*****************************************************************************/
/* Description: ***************************************************************
*****************************************************************************/
/* Analysis: ******************************************************************
*****************************************************************************/
/*****************************************************************************/
#include
#include
#include
#include
using namespace std;
#define N 100005
int x[N],y[N],mp[N];
double dp[N];
int main()
{
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
int n,m;
while(cin>>n>>m&&(n||m))
{
int i=0;
memset(mp,-1,sizeof(mp));
memset(dp,0,sizeof(dp));
while(i++>x[i]>>y[i];
mp[x[i]]=y[i];
}
for(i=n-1;i>=0;i--)
{
if(mp[i]!=-1)dp[i]=dp[mp[i]];
else
{
for(int j=1;j<=6;j++)
dp[i]+=dp[i+j];
dp[i]=dp[i]/6+1;
}
}
printf("%.4f\n",dp[0]);
}
return 0;
}
