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POJ3278——Catch That Cow

2016-08-02 14:32 288 查看

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 75564		Accepted: 23857

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

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

题目大意：人和牛在同一条直线上（看作x轴），人在n位置，牛在k位置，人每次可以移动到 n-1 或 n+1 或 2*n 位置上，求人的最小移动步数。

思路：用广搜。step数组记录步数，step[i]表示在i位置上的最小步数，vis数组为标志变量，1表示该点已经走过。若未走过，则进队，vis在该点变为1。对三种情况进行搜索，如果到达 k 位置，停止搜索，返回此时的 step ；否则继续搜索。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <queue>

using namespace std;

const int MX = 100010;
int n, k, head, next;
int vis[MX], step[MX];       //vis标记数组，  step数组存储到达该点的最小步数
queue<int> q;              //广搜需要的辅助队列

int bfs()
{
q.push(n);                //从n点开始搜
vis
= 1;
step
= 0;
while(!q.empty())
{
head = q.front();
q.pop();
for( int i = 0;i < 3;i++ )   //三种情况
{
if( !i )
{
next = head - 1;
}
else if( i==1 )
{
next = head + 1;
}
else
next = 2*head;
if( next>MX||next<0 )     //如果越界，则继续下一种情况
continue;
if( !vis[next] )
{
q.push(next);
step[next] = step[head] + 1;
vis[next] = 1;
}
if( next==k )                        //到达k位置，返回此时的步数
return step[next];
}
}
return -1;
}

int main()
{
cin>>n>>k;
memset(vis,0,sizeof(vis));
if( n>=k )                       //若n>=k  则只能一步一步的后退
cout<<n-k<<endl;
else
cout<<bfs()<<endl;

return 0;
}

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标签： poj 广搜

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