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[200005];

struct node{
int x;
int dis;

};
node u,v;

int bfs(int start,int end){
u.x=start;
u.dis=0;
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=0;i<3;i++){
if(i==0)
v.x=u.x+1;
else if(i==1)
v.x=u.x-1;
else if(i==2)
v.x=u.x*2;
if(!vis[v.x]&&v.x>=0&&v.x<=100001){
vis[v.x]=true;
v.dis=u.dis+1;
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 0;
}