poj3278 BFS入门
Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.
* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.
If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?
Input
Output
Sample Input
5 17
Sample Output
4
Hint
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;
const int maxn=4e5+5;
int n,m;
queue<pair<int,int> >q;
bool hash[maxn];
int main()
{
int m,n;
while(cin>>n>>m)
{
memset(hash,false,sizeof(hash));
pair<int,int> p;
p.first=n;
p.second=0;
hash[n]=true;
q.push(p);
while(!q.empty())
{
pair<int,int> a,b;
a=q.front();
if(a.first==m)
{
cout<<a.second<<endl;
break;
}
a.second++;
b=a;
if(b.first>m)
{
b.first=a.first-1;
if(hash[b.first]==false)
{
hash[b.first]=true;//标记该点,表示已经走过
q.push(b);
}
}
if(b.first<m)
{
b.first=a.first+1;
if(hash[b.first]==false)
{
hash[b.first]=true;//标记该点,表示已经走过
q.push(b);
}
b.first=a.first*2;
if(hash[b.first]==false)
{
hash[b.first]=true;
q.push(b);
}
b.first=a.first-1;
if(hash[b.first]==false)
{
hash[b.first]=true;
q.push(b);
}
}
q.pop();//弹出队列第一个元素
}
while(!q.empty())
q.pop();//注意清空队列
}
return 0;
}