Aeroplane chess
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1961 Accepted Submission(s): 1294
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.
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.
概率dp,參考了一些人的代碼,想明白一些問題,,
/** Author: ☆·aosaki(*’(OO)’*) niconiconi★ **/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <list>
#include <stack>
//#include <tuple>
#define mem(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define lp(k,a) for(int k=1;k<=a;k++)
#define lp0(k,a) for(int k=0;k<a;k++)
#define lpn(k,n,a) for(int k=n;k<=a;k++)
#define lpd(k,n,a) for(int k=n;k>=a;k--)
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define lowbit(x) (x&(-x))
#define ll long long
#define pi pair<int,int>
#define vi vector<int>
#define PI acos(-1.0)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define TT cout<<"*****"<<endl;
#define TTT cout<<"********"<<endl;
inline int gcd(int a,int b)
{
return a==0?b:gcd(b%a,a);
}
#define INF 1e9
#define eps 1e-8
#define mod 10007
#define MAX 10010
using namespace std;
int n,m,a,b;
int x[1000010];
double dp[1000010];
int main()
{
//freopen("in.txt","r",stdin);
while(~sc2(n,m))
{
if(n==0 && m==0) return 0;
mem(x);
lp(i,m)
{
sc2(a,b);
x[a]=b;
}
mem(dp);
dp[n]=0;
lpd(i,n-1,0)
{
if(x[i]) dp[i]=dp[x[i]];
else
{
lpn(j,1,6)
dp[i]+=dp[i+j]/6.0;
dp[i]+=1;
}
}
printf("%.4f\n",dp[0]);
}
return 0;
}