POJ 3278 Catch That Cow

POJ 3278 Catch That Cow

题目链接http://poj.org/problem?id=3278

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

题意：

给你一个人在一维坐标的坐标点，给一头牛ude坐标。对于人有三种操作，往左走，往右走，坐标值*2走法。每种操作时间加一。问最短多久catch这头牛。

题解：

用的bfs注意设置阈值。免得爆了。

有个比较迷的地方就是bfs里面的队列开在全局那么在bfs结尾没有返回值能过，或者队列开在bfs里面，在bfs函数最后随便返回一个值也能过。但是队列开在bfs里面并且结尾没有返回值则W。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
bool vis[223456];
int step[223456];
int n,k;
queue <int> q;
int bfs()
{
	
	q.push(n);
	vis[n] = 1;
	step[n] = 0;
	while (!q.empty()){
		int temp = q.front();
		q.pop();
		int temp1;
		for (int i = 1;i <= 3;i++){
			if (i == 1)
				temp1 = temp + 1;
			else if (i == 2)
				temp1 = temp - 1;
			else if (i == 3)
				temp1 = temp * 2;

			if (temp1 > 110000 || temp1 < 0)
				continue;
			if (!vis[temp1]){
				vis[temp1] = 1;
				step[temp1] = step[temp] + 1;
				q.push(temp1);
			}
			if (temp1 == k)
				return step[k];
		}
	}
}
int main(int argc, char const *argv[])
{
	while (cin>>n>>k){
		memset(vis,0,sizeof(vis));
		int ans ;
		if (n < k)
			ans = bfs();
		else ans = n - k;
		cout<<ans<<endl;
	}
	return 0;
}