Hdoj 2717.Catch That Cow 题解

Problem 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

HintThe 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

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思路

考验对状态的理解，每次走下一步有3种状态，bfs即可

代码

#include<bits/stdc++.h>
using namespace std;
const int d[3] = {1,-1,0};
struct node
{
	int pos;
	int step;
};
int n,k;
bool vis[200010];

bool judge(int x)
{
	if(vis[x] || x<0 || x>100000)
		return false;
	return true;
}
int bfs(node st)
{
	queue<node> q;
	q.push(st);
	node next,now;
	memset(vis,false,sizeof(vis));
	vis[st.pos] = true;
	while(!q.empty())
	{
		now = q.front();
		q.pop();
		if(now.pos == k) return now.step;
		for(int i=0;i<3;i++)
		{
			if(i==0 || i==1)
				next.pos = now.pos + d[i];
			else
				next.pos = now.pos * 2;
			next.step = now.step + 1;
			if(judge(next.pos))
			{
				q.push(next);
				vis[next.pos] = true;
			}
		}
	}
}

int main()
{
	while(cin>>n>>k)
	{
		node t;
		t.pos = n; t.step = 0;
		int ans = bfs(t);
		cout << ans << endl;
	}
	return 0;
}