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Catch That Cow
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 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 |
题意 告诉你两个点 n k 计算从 n到k 走的最短步数 移动方式 向n+1 or n-1 or n*2;
基本广搜水题 三个方向 搜索最短路径
#include<iostream>
#include<cstdio>
#include<cstring>
const int M=200005;
using namespace std;
typedef class{
public:
int x;
int step;
}wz;
wz q[M];
int vis[M];
int n,k;
int bfs()
{
int front=0,rear=0;
q[rear].x=n;
q[front].x=n;
q[rear++].step=0;
vis[n]=1;
while(front<rear)
{
wz w=q[front++];
if(w.x==k)
return w.step;
if(w.x-1>=0&&!vis[w.x-1])
{
vis[w.x-1]=1;
q[rear].x=w.x-1;
q[rear++].step=w.step+1;
}
if(w.x+1<=k&&!vis[w.x+1])
{
vis[w.x+1]=1;
q[rear].x=w.x+1;
q[rear++].step=w.step+1;
}
if(w.x<=k&&!vis[2*w.x])
{
vis[2*w.x]=1;
q[rear].x=w.x*2;
q[rear++].step=w.step+1;
}
}
}
int main()
{
int i;
while(cin>>n>>k)
{
memset(vis,0,sizeof vis);
cout<<bfs()<<endl;
}
return 0;
}
poj 3278 Catch That Cow（广搜）,布布扣,bubuko.com
原文地址：http://blog.csdn.net/fanerxiaoqinnian/article/details/37873637
