hdu 4405 Aeroplane chess(全期望公式)
2017-01-14 21:53
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Aeroplane chess
[b]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3785 Accepted Submission(s): 2404
[/b]
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
2 0
8 3
2 4
4 5
7 8
0 0
[align=left]Sample Output[/align]
1.1667
2.3441
#include<bits/stdc++.h>
using namespace std;
#define MAXN 100010
int t[MAXN];
double dp[MAXN];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)&&(n||m))
{
memset(t,0,sizeof(t));
for(int i=0;i<m;i++)
{
int x,y;
cin>>x>>y;
t[x]=y;
}
for(int i=0;i<6;i++)
dp[n+i]=0;
for(int i=n-1;i>=0;i--)
{
if(t[i])
dp[i]=dp[t[i]];
else
{
double mid=0;
for(int j=1;j<=6;j++)
mid+=dp[i+j];
dp[i]=mid/6+1;
}
}
printf("%.4f\n",dp[0]);
}
return 0;
}
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