POJ 3278:Catch That Cow 抓住那头牛

POJ 3278:Catch That Cow 抓住那头牛

总时间限制: 

2000ms 

内存限制: 

65536kB

描述

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?

输入

Line 1: Two space-separated integers:  N and  K

输出

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

样例输入

5 17

样例输出

4

提示

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.

题目分析：

Finding the shortest path，采用广度搜索BFS的方法。

注意分析题目：

Farmer John 的位置是N， cow 牛的位置是 K；

action只有3种，+1，-1，*2； 没有 “除以2”。

所以当N>K，时，只能执行-1 action。 即 如N-K，最短steps= K-N。

当N

N>0; 可以执行-1 action。

N

2*N

样例代码：

/*
广度优先搜索，重要的是剪枝和标记访问过的路径。
所谓的剪枝，也可以理解为生成每一个子节点的条件；所谓条件限制越严谨，剪枝效果越好。

*/
#include 
#include 
#include 

using namespace std;

const int MAX=100020; // 这个MAX的设定并不太严谨
int visited[MAX]={0};

int main()
{
	int n,k;
	while (cin>>n>>k){
		if( n > k ){ 
			cout< q;
		q.push(n);
		int t=0;
		while( !q.empty() ){
			t = q.front();
			q.pop();

			if(t == k){
				break;
			}
			if( t < k && !visited[t+1] ){ // 当t>k时，不执行+1 action
				q.push(t+1); 
				visited[t+1] = visited[t] + 1;
			}

			if( t > 0 && !visited[t-1] ){ //
				q.push(t-1);
				visited[t-1] = visited[t] + 1;
			}
			
			if( 2*t < MAX && !visited[2*t] ){ //此处不太严谨，如上讨论
				q.push(2*t);
				visited[2*t] = visited[t] +1;
			}
		}
		cout<