poj 3278 Catch That Cow BFS

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 77719		Accepted: 24552

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.

题意：

有一个农民和一头牛，他们在一个数轴上，牛在k位置保持不动，农户开始时在n位置。设农户当前在M位置，每次移动时有三种选择：1.移动到M-1；2.移动到M+1位置；3.移动到M*2的位置。问最少移动多少次可以移动到牛所在的位置。所以可以用广搜来搜索这三个状态，直到搜索到牛所在的位置。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int ans = 0;
bool vis[233333];
struct qer
{
	int x, step;
};
queue <qer> q;
void bfs(int x, int s)
{
	q.push((qer){x, 0});
	vis[x] = 1;
	while(!q.empty())
	{
		qer ls = q.front();
		q.pop();
		if(ls.x == s)
		{
			cout<<ls.step;
			return ;
		}
		if(ls.x >= 1 && !vis[ls.x - 1])
        {
            qer temp;
            vis[ls.x - 1] = 1;
            temp.x = ls.x - 1;
            temp.step = ls.step + 1;
            q.push(temp);
        }
        if(ls.x <= s && !vis[ls.x + 1])
        {
        	qer temp;
        	vis[ls.x + 1] = 1;
        	temp.x = ls.x + 1;
        	temp.step = ls.step + 1;
        	q.push(temp);
        }
        if(ls.x <= s && !vis[ls.x * 2])
        {
        	qer temp;
        	vis[ls.x * 2] = 1;
        	temp.x = ls.x * 2;
        	temp.step = ls.step + 1;
        	q.push(temp);
        }
	}
}
int main()
{
	int n, k;
	cin>>n>>k;
	bfs(n, k);
	return 0;
}