Catch That Cow（简单BFS)

Link：http://poj.org/problem?id=3278

Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 51720   Accepted: 16230 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 USACO 2007 Open Silver

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AC  code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;
struct node{
	int sp;
	int num;
}st;
int n,k;
bool vis[100001];
int BFS()
{
	queue<node>q;
	node cur,nex;
	memset(vis,false,sizeof(vis));
	st.sp=0;
	st.num=n;
	q.push(st);
	vis[st.num]=true;
	while(!q.empty())
	{
		cur=q.front();
		q.pop();
		if(cur.num==k)
		{
			return cur.sp;
		}
		if(cur.num-1>=0&&!vis[cur.num-1])
		{
			nex.num=cur.num-1;
			nex.sp=cur.sp+1;
			vis[nex.num]=true;
			q.push(nex);
		}
		if(cur.num+1<=100000&&!vis[cur.num+1])
		{
			nex.num=cur.num+1;
			nex.sp=cur.sp+1;
			vis[nex.num]=true;
			q.push(nex);
		}
		if(cur.num*2<=100000&&!vis[cur.num*2])
		{
			nex.num=cur.num*2;
			nex.sp=cur.sp+1;
			vis[nex.num]=true;
			q.push(nex);
		}
	}
}
int main()
{
	int ans;
	while(cin>>n>>k)
	{
		ans=BFS();
		cout<<ans<<endl;
	}
	return 0;
}