POJ 题目Catch That Cow（BFS）

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 52537 Accepted: 16471

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

ac代码

#include<stdio.h>#include<string.h>#include<queue>#include<iostream>#define INF 0xfffffff#define min(a,b) (a>b?b:a)using namespace std;struct s{int pos,step;}a,temp;int n,k,ans,vis[100100];void bfs(){memset(vis,0,sizeof(vis));a.pos=n;a.step=0;queue<struct s>q;q.push(a);vis[a.pos]=1;while(!q.empty()){a=q.front();q.pop();for(int i=0;i<3;i++){if(i==0)temp.pos=a.pos+1;if(i==1)temp.pos=a.pos-1;if(i==2)temp.pos=a.pos*2;temp.step=a.step+1;if(temp.pos==k){ans=min(ans,temp.step);continue;}if(temp.pos<0||temp.pos>100001)continue;if(!vis[temp.pos]){vis[temp.pos]=1;q.push(temp);}}}}int main(){while(scanf("%d%d",&n,&k)!=EOF){ans=INF;if(n==k){printf("0\n");continue;}if(n>k){printf("%d\n",n-k);continue;}bfs();printf("%d\n",ans);}}