HDU 2717 Catch That Cow(哎！居然没想到用bfs)

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9842 Accepted Submission(s): 3079

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 

/*终于AC了  */
//看了别人的代码 其实可以用结构体  建个数组记忆是否进过队  并且N不用那么大
//N=100000+10就够了 因为太大也不可能是答案
//优化空间可以改成  N=100000+10   
//（不知道每次查长度需要时间不） 要的话改成结构体可以省时间 
#include <iostream>
#include <queue>
#include <cstdio>
#include <string>
#include <string.h>
using namespace std;
const int N=100000*6+10;
queue<int> q;
bool book[N];
int main(){
	int a,b,re,tmp,i;
	while(scanf("%d%d",&a,&b)==2){
		while(!q.empty()) q.pop();
		re=0;
		memset(book,0,sizeof(book));
		book[a]=1;
		 q.push(a);
		 while(!q.empty()){
		 	int len=q.size();
		 	for(i=0;i<len;i++){
		 	   tmp=q.front();
		 	   	q.pop();	
		 	   if(tmp==b)
		 	     goto AC;
				  if(!book[tmp+1]){
				     book[tmp+1]=1;
				  		q.push(tmp+1);
				  }   
			 if(tmp>0&&!book[tmp-1]){
				     book[tmp-1]=1;
				  	q.push(tmp-1);
				  }
			if(tmp*2<N&&!book[tmp*2]){
				     book[tmp*2]=1;
				  	q.push(tmp*2);
				  }	 	 		
			 }	 
			 re++;	 	
		 }
		 AC:;
		  printf("%d\n",re);
	}
	return 0;
}


/*！！爆内存
#include <iostream>
#include <queue>
#include <cstdio>
using namespace std;
const int N=100000;
struct Node{
	int v,t;
}node,tmp;
queue<Node> q;
int main(){
	int a,b;
	while(scanf("%d%d",&a,&b)==2){
		while(!q.empty()) q.pop();
		 node.v=a;
		 node.t=0;
		 q.push(node);
		 while(!q.empty()){
		 	node=q.front();
		 	q.pop();
		 	if(node.v==b)
		 	break;
		 	tmp.v=node.v+1;
		 	tmp.t=node.t+1;
		 	q.push(tmp);
		 	tmp.v=node.v-1;
		 	tmp.t=node.t+1;
		 	q.push(tmp);
		 	tmp.v=node.v*2;
		 	tmp.t=node.t+1;
		 	q.push(tmp);
		 }
		  printf("%d\n",node.t);
	}
	return 0;
}
*/

/*想水一下 水不过  TLE
#include <iostream>
#include <cstdio>
using namespace std;
const int N=100000;
int b,re;
void dfs(int now,int sum){
	if(sum>re)
	return;
	else if(now==b){
		if(sum<re)
		  re=sum;
		  return;
	}
	dfs(now+1,sum+1);
	dfs(now-1,sum+1);
	dfs(now*2,sum+1);
}
int main(){
	int a;
	while(scanf("%d%d",&a,&b)==2){
		  re=0;
		  int tmp=a;
		  while(tmp<b){
		  	  tmp*=2;
		  	  re++;
		  }
		  re+=tmp-b;
		  dfs(a,0);
		  printf("%d\n",re);
	}
	return 0;
}
*/