poj 3278 Catch That Cow bfs

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 106185		Accepted: 33180

DescriptionFarmer 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?
InputLine 1: Two space-separated integers: N and KOutputLine 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.Sample Input5 17Sample Output4HintThe fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.
题意：给你n和k
n有三种变化方式
 n=n+1; n = n-1; n = n*2；
问你经过几次变化 n==k；

思路：这个题是归类到了bfs中 很好想 三方向的bfs
如果是直接碰到的话 可能想不出来.....
不过写完之后就re了...
坑点 n在变化的时候  n-- 和 n *= 2 的时候会发生数组越界 从而re 所以需要剪枝

ac代码#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>

using namespace std;

int step[100005];
bool vis[100005];

queue<int>q;

int bfs(int n,int k)
{
int head,next;
q.push(n);
vis
= true;
step
= 0;
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)
{
if(head > 0)//剪枝
next = head - 1;
else
continue;
}
else if( i == 2)
{
if(head <= 50000)//剪枝
next = head*2;
else
continue;
}

if(!vis[next])
{
q.push(next);
step[next] = step[head] + 1;
vis[next] = true;
}
if(next == k)

a20b
return step[next];
}

}
}

int main()
{
int n,k;
while(~scanf("%d %d",&n,&k))
{
memset(vis,false,sizeof(vis));
memset(step,0,sizeof(step));

while(!q.empty())
q.pop();

if(n >= k)
printf("%d\n",n-k);
else
printf("%d\n",bfs(n,k));
}

return 0;
}