Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

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

USACO 2007 Open Silver

bfs注意注释的几个地方，坑哭我了，，，只怪自己太菜

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int vis[110000];
int n,k;
struct node{
	int val,d;
}s,num;

int bfs(){
	memset(vis,0,sizeof(vis));
	queue <node> Q;
	while(!Q.empty()) Q.pop();
	s.val=n; s.d=0;
	vis[n]=1;
	Q.push(s);
	while(!Q.empty()){
		s=Q.front();
		Q.pop();
		for(int i=0;i<3;i++){
			if(i==0) num.val=s.val+1;
			if(i==1) num.val=s.val-1;
			if(i==2) num.val=s.val*2;
			num.d=s.d+1;
			if(num.val==k) return num.d;
			if(num.val<0 || num.val>110000)//原来写的num.val>k，wa了20多次，草，fuck，，，
				continue;
			//if(num.val>=0&&!vis[num.val]&&num.val<100000){//这样写num.val大于110000是，vis数组就会溢出
			if(!vis[num.val]){
				vis[num.val]=1;
				Q.push(num);
			}
 		}
	}
	return -1;
}
int main(){
	while(scanf("%d%d",&n,&k)!=EOF){
		if(n>k) printf("%d\n",n-k);
		else if(n==k) printf("0\n");
		else
			printf("%d\n",bfs());
	}
	return 0;
}