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Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX
+ 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?
5 17
4
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
int n,m;
int vis[200000];
struct node{
int x,y,step;
}pp;
queue<node>q;
int bfs()
{
while(!q.empty())
{
q.pop();
}
q.push(pp);
memset(vis,0,sizeof(vis));
vis[pp.x]=1;
while(!q.empty())
{
node gg=q.front();
if(gg.x==m)
return gg.step;
q.pop();
for(int i=0;i<3;i++)
{
node dd=gg;
if(i==0)
dd.x=dd.x+1;
else if(i==1)
dd.x=dd.x-1;
else if(i==2)
dd.x=dd.x*2;
dd.step++;
if(dd.x==m)
return dd.step;
if(dd.x>=0&&dd.x<=200000&&!vis[dd.x])
{
vis[dd.x]=1;
q.push(dd);
}
}
}
return 0;
// memset(bis,0,sizeof(vis));
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
pp.x=n;
pp.step=0;
int time=bfs();
printf("%d\n",time);
}
return 0;
}
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hdoj 2717 Catch That Cow 【bfs】
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原文地址：http://blog.csdn.net/qq_21654717/article/details/47319369
