HDOJ-三部曲一(搜索、数学)-1006- Catch That Cow

Catch That Cow

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 48   Accepted Submission(s) : 16

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

 

Source

PKU

 

 

这道题是到BFS水题，不过我在这道题上re了很多次。。。原因是我没有剪枝。

 

 

 

#include<iostream>
#include<cstring>
using namespace std;
int que[1000001];                        //数组要开大谢；大数组用全局变量开，不能再函数里开。
int step[1000001]={0};   
bool f[1000001]={false};
int n,k;

int BFS()
{
	int front=0,rear=1;
	que[0]=n;
	step[0]=0;
	f[n]=true;
	if(que[front]==k)
		return step[front];
	while(front<rear)
	{
		if(que[front]<k&&!f[que[front]+1])   //只有当前的数比k小时，加一才能更接近k。
		{
			que[rear]=que[front]+1;
			step[rear]=step[front]+1;
			f[que[rear]]=true;
			if(que[rear]==k)
				return step[rear];
			rear++;
		}
		if(que[front]-1>=0&&!f[que[front]-1])  //数组不能越界
		{
			que[rear]=que[front]-1;
			step[rear]=step[front]+1;
			f[que[rear]]=true;
			if(que[rear]==k)
				return step[rear];
			rear++;
		}
		if(que[front]<k&&!f[que[front]*2])  //只有当前的数比k小时，乘2才有可能更接近k。
		{
			que[rear]=que[front]*2;
			step[rear]=step[front]+1;			
			f[que[rear]]=true;
			if(que[rear]==k)
				return step[rear];
			rear++;
		}
		front++;
	}
}

int main()
{
	while(cin>>n>>k)
	{
		memset(que,0,sizeof(que));
		memset(step,0,sizeof(step));
		memset(f,false,sizeof(f));
		cout<<BFS()<<endl;	
	}
}