POJ 3278 Catch That Cow 【bfs+dp】

追捕逃逸奶牛的算法

原创于 2019-04-07 21:38:56 发布 · 229 阅读

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本博客介绍了一个寻找逃逸奶牛的算法问题，通过行走与瞬移两种方式，利用广度优先搜索（BFS）算法找到从起点到目标点的最短路径。示例展示了从起始位置5到达目标位置17所需的最少时间。

Time Limit: 2000MS

Memory Limit: 65536K
Total Submissions: 32679

Accepted: 10060

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.

#include<iostream>
#include<string.h>
using namespace std;
int dp[100010];
int q[100010];
int n,k;
int a[3]={1,-1,2};
int bfs(int nn){
	dp[nn]=1;
	int p=0,r=0;
	q[p]=nn;
	while(p<=r){
		int ss=q[p];
		for(int i=0;i<=2;i++){
			int idd;
			if(i==2){
				idd=ss*a[i];
			}else{
				idd=ss+a[i];
			}
			if(!dp[idd]&&idd>=1&&idd<=100000){
				r++;
				q[r]=idd;
				dp[idd]=dp[ss]+1;
				if(idd==k)
			       return dp[idd]-1;
			}	
		}
		p++;
	}
}
int main(){
	memset(dp,0,sizeof(dp));
	scanf("%d%d",&n,&k);
	if(n>=k){
		cout<<n-k<<endl;
	}else{
		cout<<bfs(n)<<endl;
	}
	return 0;	
}