hdu 2717 Catch That Cow

Problem 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

HintThe 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.

简单广搜，坑爹的注意起点等于终点就行了；
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;

struct node{
int x,step;
}s,e,q,p;
int dir[8][2]={{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1},{-2,1},{-1,2}};
int next1[2]={1,-1};
int vis[1000005];
bool judge(node a){
if(a.x<0||a.x>1000000||vis[a.x]) return false;
return true;
}
int bfs(){
queue<node>que;
s.step=0;
vis[s.x]=1;
que.push(s);int s=1;
while(!que.empty()){
q=que.front();//printf("%d->%d\n",q.x,q.step);
que.pop();

for(int i=0;i<2;i++){
p=q;
p.x=q.x+next1[i];
if(judge(p)){
vis[p.x]=1;
p.step++;
que.push(p);//printf("%d %d %d %d\n",p.x,p.y,e.x,e.y);
if(p.x==e.x)
return p.step;
}
}
p=q;
p.x=q.x*2;//printf("%d~~\n",p.x);
if(judge(p)){
vis[p.x]=1;
p.step++;
que.push(p);
if(p.x==e.x)
return p.step;
}
}
}
int main(){
int a,b;
while(~scanf("%d%d",&a,&b)){
memset(vis,0,sizeof(vis));
s.x=a,e.x=b;
if(s.x==e.x)printf("0\n");
else
printf("%d\n",bfs());
}
return 0;
}