POJ 3278 Catch That Cow

【分析】简单bfs，来不及解释了，快上车。

【AC代码】#include<iostream>
#include<queue>
#include<string.h>
#include<stdio.h>
using namespace std;
const int maxn=100001;
bool vis[maxn];
int step[maxn];
queue <int> q;
int bfs(int n,int k)
{
int head,next;
q.push(n);
step
=0;
vis
=true;
while(!q.empty())
{
head=q.front();
q.pop();
for(int i=0;i<3;i++)
{
if(i==0) next=head-1;
else if(i==1) next=head+1;
else next=head*2;
if(next<0 || next>=maxn) continue;
if(!vis[next])
{
q.push(next);
step[next]=step[head]+1;
vis[next]=true;
}
if(next==k)
return step[next];
}
}
}
int main()
{
int n,k;
scanf("%d %d",&n,&k);
memset(step,0,sizeof(step));
memset(vis,false,sizeof(vis));
while(!q.empty()) q.pop();
if(n>=k) printf("%d\n",n-k);
else printf("%d\n",bfs(n,k));
return 0;
}