poj 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 43187 Accepted: 13435

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.

bfs+优先队列

#include"stdio.h"#include"string.h"#include"queue"using namespace std;#define N 100005int mark[N],k,n;struct node{int x,step;friend bool operator<(node a,node b){return a.step>b.step ;       //用时少的优先级高}};int judge(int x){if(x>=0&&(x<=n||x<k+2)&&!mark[x])         //N可能远大于K，故x<=k+2不对哦!!return 1;return 0;}int bfs(int x){priority_queue<node>q;node cur,next;cur.x=x;cur.step=0;q.push(cur);mark[x]=1;while(!q.empty()){cur=q.top();if(cur.x==k)return cur.step;q.pop();next.step=cur.step+1;       next.x=cur.x+1;//+1if(judge(next.x)){mark[next.x]=1;q.push(next);}next.x=cur.x-1; //-1if(judge(next.x)){mark[next.x]=1;q.push(next);}next.x=cur.x*2;   //*2if(judge(next.x)){mark[next.x]=1;q.push(next);}}return 0;}int main(){while(scanf("%d%d",&n,&k)!=-1){memset(mark,0,sizeof(mark));printf("%d\n",bfs(n));}return 0;}