POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 80477		Accepted: 25346

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

Line 1: Two space-separated integers: N and K
Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
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;
while(cin>>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;
}

这是一个比较简单的bfs搜索算法的算法题，这段代码最重要的是

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;
}

在for语句中使用了一个queue，不断地将步数（step）相同的值放入其中

比如：第一次放入queue的为n，后又清空

            第二次放入的为n-1，n+1，2*n，而这第二次的三个next值的步数都为1，这个很重要，对于bfs而言，就是需要找到步数一致的树的一个结点

自己好弱呀，看了好久才懂