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POJ 【3278】 Catch That Cow

2016-04-15 08:15 387 查看

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 70204		Accepted: 22082

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

/*
*这道题有几个坑点需要注意，首先我们可以判断出负坐标是没有作用的
*判断数组需要开2倍大，第二数组处理过程中不能越界.
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
using namespace std;
const int maxn = 100005;
int vis[2*maxn];
int n,k;
struct Node
{
int x,t;
Node(int a,int b):x(a),t(b){}
Node(){}
};

bool judge(int loc)
{
if(loc < 0 || loc >= 2*maxn)
return false;
return true;
}

void bfs()
{
queue <Node> que;
while(!que.empty()) que.pop();
Node cur;
memset(vis,0,sizeof(vis));
que.push(Node(n,0));
vis
= 1;
while(!que.empty())
{
cur = que.front();
que.pop();
if(cur.x == k)
{
printf("%d\n",cur.t);
return ;
}
if(judge(cur.x+1) && !vis[cur.x+1]) // 注意judge放在判断vis数组前面，否则会runtime error
{
vis[cur.x+1] = 1;
que.push(Node(cur.x+1,cur.t+1));
}
if(judge(cur.x-1) && !vis[cur.x-1])
{
vis[cur.x-1] = 1;
que.push(Node(cur.x-1,cur.t+1));
}
if(judge(cur.x*2) && !vis[cur.x*2])
{
vis[cur.x*2] = 1;
que.push(Node(cur.x*2,cur.t+1));
}
}
}

int main()
{
while(scanf("%d%d", &n, &k) != EOF)
bfs();
return 0;
}

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