Catch That Cow(广度优先搜索_bfs)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 48036		Accepted: 15057

DescriptionFarmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (0 ≤ N ≤ 100,000) on a number line and the cow is at a pointK (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 orX + 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?InputLine 1: Two space-separated integers: N andKOutputLine 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

题意：输入两个数n，k。求从n到k最少走多少步。可以前进1后退1或者当前的位置*2；

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>using namespace std;struct node{int x;//当前位置int ans;//走的步数}q[1000010];int vis[1000010];//标记变量，该点是否被访问；int jx[]={-1,1};//后退1或者前进1；struct node t,f;int n,k;void bfs(){int i;int s=0,e=0;//指针模拟队列。往队列加e++ 往队列里提出数s++memset(vis,0,sizeof(vis));t.x=n;//当前初始位置vis[t.x]=1;//标记为1代表访问过；t.ans=0;//初始位置步数为0；q[e++]=t;//把当前步数加人队列while(s<e)//当队列不为空{t=q[s++];//提出if(t.x==k)//如果该数正好等于目标位置直接输出步数{printf("%d\n",t.ans);break;}for(i=0;i<3;i++)//i=0后退一步，i=1前进一步，i=2此时的位置*2；{if(i==2){f.x=t.x*2;}else{f.x=t.x+jx[i];}if(f.x>=0&&f.x<=100000&&!vis[f.x]){f.ans=t.ans+1;q[e++]=f;vis[f.x]=1;}}}}int main(){while(~scanf("%d %d",&n,&k)){bfs();}return 0;}