POJ 3278-Catch That Cow 广度优先搜索BFS

题目来源：http://acm.pku.edu.cn/JudgeOnline/problem?id=3278



解题报告：



广度优先搜索，对点X，它邻接的点有X-1, X+1, 2×X

这样从N开始对N邻接的点进行搜索，然后再以此广度搜索下去，直到搜索到K

得到的从N到K的最短路径就是 最短时间。

#include <iostream>
#include <queue>
using namespace std;

int visit[200001];
int d[200001];

int BFS(int N, int K)
{
	int s=N;
	visit[s]=1; //灰色
	d[s]=0;
	queue<int> Q;
	Q.push(s);
	while(!Q.empty())
	{
		int u=Q.front();
		Q.pop();
		int v[3];
		v[0]=u-1;
		v[1]=u+1;
		v[2]=2*u;
		for(int i=0;i<3;i++)
		{
			if(v[i]>=0 && v[i]<=200001)
			{
				if(visit[v[i]]==0)
				{
					visit[v[i]]=1;
					d[v[i]]=d[u]+1;
					if(v[i]==K)
						return d[K];
					Q.push(v[i]);
				}
			}
		}
		visit[u]=2;
	}
	return 0;
}

int main()
{
	int N,K;
	cin >> N >> K;
	cout << BFS(N,K) << endl;

}

附录：



Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 18412		Accepted: 5657

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

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