poj 3278Catch That Cow(hd2717)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 61286		Accepted: 19141

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.
Sample Input

5 17

Sample Output

4

Hint
The 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.
嗯，题目大一就是说，给n和k，n代表人所在位置，k代表cow 所在位置，人要抓住牛，但人只能一次只能走到n-1或n+1或2*n位置，每走一次花费一分钟，问最快多久能抓到牛。开始写的时候老wa后来看讨论区特殊测试数据2 2和0 10000和10000 0 最后一个不出结果，改一下才过

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int mark[100010];
int step[100010];
void bfs(int s,int e)
{
int ss,nn;
mark[s]=1;
step[s]=0;
queue<int>q;
q.push(s);
while(!q.empty())
{
ss=q.front();
q.pop();
if(ss==e)
{
printf("%d\n",step[ss]);
return ;
}
nn=ss+1;
if(nn<=100001&&!mark[nn])//就是每次这里比较的时候开始写的nn<=e
{
mark[nn]=1;
step[nn]=step[ss]+1;
q.push(nn);
}
nn=ss-1;
if(nn>=0&&nn<=100001&&!mark[nn])
{
mark[nn]=1;
step[nn]=step[ss]+1;
q.push(nn);
}
nn=ss*2;
if(nn<=100001&&!mark[nn])
{
mark[nn]=1;
step[nn]=step[ss]+1;
q.push(nn);
}
}
}
int main()
{
int n,k;
while(~scanf("%d%d",&n,&k))
{
memset(mark,0,sizeof(mark));
memset(step,0,sizeof(step));
bfs(n,k);
}
return 0;
}