poj3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36079		Accepted: 11123

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.

完全是BFS，沒有什麼技巧，主要還是用VISIT標記，防重搜

#include<iostream>#include<stdio.h>#include<cstring>using namespace std; struct tree{int x;int step;} queue[800050];int visit[100050];int n,k;int bfs(){int t,w,temp,minx=10000000,xx;t=w=1;queue[t].x=n;visit[n]=1;queue[t].step=0;while(t<=w){ xx=queue[t].x;if(xx>k){temp=queue[t].step+xx-k;//如果比結果大，只能減，所以可以直接求出來！if(temp<minx)minx=temp;t++;//這裡的T一定要加，要不然就錯了continue;}if(xx*2<=100000&&(!visit[2*xx])){queue[++w].x=xx*2;visit[2*xx]=1;//用VISIT來標記，防止重搜queue[w].step=queue[t].step+1;if(queue[w].x==k){if(queue[w].step<minx)return queue[w].step;else return minx;}}if(xx+1<=100000&&(!visit[xx+1])){queue[++w].x=xx+1;visit[xx+1]=1;queue[w].step=queue[t].step+1;if(queue[w].x==k){if(queue[w].step<minx)return queue[w].step;else return minx;}}if(xx>=1&&(!visit[xx-1])){queue[++w].x=xx-1;visit[xx-1]=1;queue[w].step=queue[t].step+1;if(queue[w].x==k){if(queue[w].step<minx)return queue[w].step;else return minx;}}t++;}return -1;}int main (){while(scanf("%d%d",&n,&k)!=EOF){memset(visit,0,sizeof(visit));if(n>=k)printf("%d\n",n-k);elseprintf("%d\n",bfs());}return 0;}