HDOJ 题目2717 Catch That Cow（BFS）

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7202    Accepted Submission(s): 2274

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

 

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#include<stdio.h>#include<iostream>#include<string.h>#include<queue>using namespace std;int n,k;int v[100010],d[100010];void bfs(){queue<int>q;q.push(n);while(!q.empty()){int u=q.front();if(u==k)return;q.pop();int next=u-1;if(next>=0&&next<1000001&&!d[next]){q.push(next);d[next]=d[u]+1;v[next]=1;}next=u+1;if(next>=0&&next<100001&&!d[next]){q.push(next);d[next]=d[u]+1;v[next]=1;}next=u*2;if(next>=0&&next<100001&&!d[next]){q.push(next);d[next]=d[u]+1;v[next]=1;}}}int main(){while(scanf("%d%d",&n,&k)!=EOF){//queue<int>q;memset(v,0,sizeof(v));memset(d,0,sizeof(d));v[n]=1;//q.push(n);bfs();printf("%d\n",d[k]);}}