poj 3278 Catch That Cow（BFS，剪枝）

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 43761		Accepted: 13650

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.
【注意】
剪枝
测试数据有多组while（cin>>n>>k）

【解题思路】
这道题目不知道错了多少次。
第一次错误是因为没有剪枝，纯纯的BFS造成内存超出。
后面就是无穷无尽的Runtime error ，是因为数组没有开的足够大，动态数组真心不行，开到100*k+100的大小还是提示RE，真心给跪了。所以还是老老实实开大数组吧。
还有就是题目中说输入只有一行，可是我把while（cin>>n>>k）去掉之后提示wrong answer

三岔路的BFS，但是要剪枝，我做的剪枝方案就是所有的数都要在2*k以内，不要太大了，没有意义。所有的数只允许在队列里出现一次，不然加一减一造成很多浪费

还有就是bool值的new G++编辑器默认给的初始值为false，但是初始化是没人人需要养成的好习惯

【code】

我自己的渣代码

#include<iostream>
#include<queue>
using namespace std;
int main(void){

	int n,k;
	while(cin>>n>>k){

	queue<int> q;
	queue<int> c;
	if(n==k){
		cout<<'0'<<endl;
		continue;
	}
	bool *in = new bool[1000000];
	for(int i =0;i<1000000;i++)in[i]=false;
	q.push(n);
	in[n]= true;
	c.push(0);
	while(!q.empty()){
		//cout<<"here "<<endl;
		int temp = q.front();
		int tempc = c.front()+1;
		q.pop();
		c.pop();
		if(temp+1==k||temp-1==k||temp*2==k){
			cout<<tempc<<endl;
			break;
		}
		if(temp+1>=0 &&in[temp+1]==false && temp+1<=2*k){
			q.push(temp+1);
			c.push(tempc);
			in[temp+1]=true;
		}
		if(temp-1>=0 && in[temp-1]==false){
			q.push(temp-1);
			c.push(tempc);
			in[temp-1]=true;
		}
		if(temp*2>=0 && temp*2<=k*2 && in[temp*2]==false){
			q.push(temp*2);
			c.push(tempc);
			in[temp*2]=true;
		}
	}
	delete[]in;
	}
	return 0;	
}

想找一个效率比较高的代码给大家参考，可惜。。。没找到，大家将就着看吧