poj 3278 Catch That Cow （BFS+剪枝）

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 58474		Accepted: 18173

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

题目链接：http://poj.org/problem?id=3278

题目大意：数字n到k，每次只能通过三种变换中的一种，即+1，-1或*2，求最少变换次数。

解题思路：BFS，剪枝特别重要，需要visit数组标记，已变换过的数字不再加入队列，能极大程度地提高效率。数组开大概200000的范围，超过次范围的数字舍去。一开始以为会出现负数，但仔细想想，因为n和k都是非负整数，变换过程如果出现负数不会是最优解，因此标记时不必考虑下标为负的情况。

代码如下：

#include <cstdio>
#include <queue>
using namespace std;
const int maxn=200010;
int m,n;
int vis1[maxn];
struct node
{
int x,step;
};
int bfs()
{
node sd;
sd.x=n;
sd.step=0;
queue<node>q;
q.push(sd);
while(!q.empty())
{
node nd=q.front();
if(nd.x==m)
return nd.step;
q.pop();
for(int i=0;i<3;i++)//枚举三种变换方法
{
if(i==0)
{
sd.x=nd.x-1;
sd.step=nd.step+1;
if(sd.x>=0&&!vis1[sd.x])  //该数字没有访问过，可加入队列
{
vis1[sd.x]=1;
q.push(sd);
}
}
else if(i==1)
{
sd.x=nd.x+1;
sd.step=nd.step+1;
if(sd.x>maxn)         //超出最大范围的数字舍去
continue;
if(sd.x>=0&&!vis1[sd.x])
{
vis1[sd.x]=1;
q.push(sd);
}
}
else
{
sd.x=nd.x*2;
sd.step=nd.step+1;
if(sd.x>maxn)
continue;
if(sd.x>=0&&!vis1[sd.x])
{
vis1[sd.x]=1;
q.push(sd);
}
}
}
}

}
int main()
{
scanf("%d%d",&n,&m);
printf("%d\n",bfs() );
return 0;
}