POJ 3278 Catch That Cow(基础题）



Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 46307 Accepted: 14516

Description

Farmer 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 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 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?

Input

Line 1: Two space-separated integers: N andK

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.
思路：用BFS,有3个方向入队列
#include <cstdio>#include <map>#include <algorithm>#include <cstring>#include <iostream>#include <cmath>#include <queue>#include <string>using namespace std;const int maxn=2000000;int dis[maxn],vis[maxn];;int st,ed;queue<int>q;void bfs(int c){q.push(c);vis[c]=1;while(!q.empty()){int temp=q.front();q.pop();if(temp==ed){return ;}if(temp+1<=maxn&&!vis[temp+1]){vis[temp+1]=1;dis[temp+1]=dis[temp]+1;q.push(temp+1);}if(temp-1>=0&&!vis[temp-1]){vis[temp-1]=1;dis[temp-1]=dis[temp]+1;q.push(temp-1);}if(temp*2<=maxn&&!vis[temp*2]){vis[temp*2]=1;dis[temp*2]=dis[temp]+1;q.push(temp*2);}}}int main(){while(scanf("%d %d",&st,&ed)!=EOF){memset(vis,0,sizeof(vis));dis[st]=0;while(!q.empty())q.pop();bfs(st);printf("%d\n",dis[ed]);}return 0;}