poj 3278 catch that cow BFS(基础水)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 61826		Accepted: 19329

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

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#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
 vis[];

struct node{
  int x;
  int dis;

};
node u,v;

int bfs(int start,int end){
     u.x=start;
     u.dis=;
     queue<node>q;
     vis[start]=true;
     q.push(u);
     while(!q.empty()){
        u=q.front();
        q.pop();
        if(u.x==end)
            return u.dis;
        for(int i=;i<;i++){
            if(i==)
                v.x=u.x+;
            else if(i==)
                v.x=u.x-;
            else if(i==)
                v.x=u.x*;
            if(!vis[v.x]&&v.x>=&&v.x<=){
                vis[v.x]=true;
                v.dis=u.dis+;
                q.push(v);
            }
        }

     }
}

int main(){
    int start,end;
    while(scanf("%d%d",&start,&end)!=EOF){
        memset(vis,false,sizeof(vis));
        int step=bfs(start,end);
        printf("%d\n",step);
    }
    return ;
}