Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 114963 | Accepted: 35950 |
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.
Source
思路：很明显的bfs，但是dfs一直卡，正在思考为什么...
代码如下：
#include<iostream>
#include<cstdio>
#include<string.h>
#include<queue>
using namespace std;
struct node{
int x,y,step;
}pp;
queue<node> q;
int vis[200010];
int n,m;
int bfs()
{
while(!q.empty()) q.pop();
memset(vis,0,sizeof(vis));
vis[pp.x]=1;//标记牧场主开始的位置
q.push(pp);
while(!q.empty()){
node now=q.front();
if(now.x==m) return now.step;
q.pop();
for(int i=0;i<3;i++){
node next=now;
if(i==0){
next.x-=1;
}
else if(i==1){
next.x+=1;
}
else if(i==2){
next.x*=2;
}
next.step++;
if(next.x==m) return next.step;
if(next.x>=0&&next.x<=200000&&!vis[next.x]){//判断是否越界
vis[next.x]=1;
q.push(next);
}
}
}
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF){
bfs();
pp.x=n,pp.step=0;
printf("%d\n",bfs());
}
}