poj 3278 Catch That Cow

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.

【题意】

给你两个数字，你需要在有限步骤之内将第一个数字变成第二个数字，其中你可以进行三种操作：

1 将第一个数字进行加一。

2 将第一个数字进行减一。

3 将第一个数字翻倍。

问你最少要经过多少次能将第一个数字变成第二个数字。比如说测栗，5(翻倍)->10(减一)->9(翻倍)->18(减一)->17。一共四步，所以输出4。

用广搜可以做出来，每个状态最多有三个子状态，逐层搜索就行。

【代码】

#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;
struct P
{
int a,b;
P(int A=0,int B=0):a(A),b(B) {}
} p1(0,0);
bool vis[400005];//因为有×2操作可能会×爆，所以数组开大一点
int bfs(int a,int b)
{
memset(vis,false,sizeof(vis));//每个数据都没有走过
vis[a]=true;
queue<P>ans;
while(!ans.empty())
ans.pop();//首先得清空队列
p1.a=a,p1.b=0;
ans.push(p1);
while(!ans.empty())
{
p1=ans.front();
ans.pop();
if(p1.a==b)//如果找到答案了，就直接返回
return p1.b;
else if(p1.a<0)//小于0的肯定不是答案
continue;
else
{
if(!vis[p1.a+1]&&p1.a<=100000)//不能越界，并且没有搜索过这个数据
{
ans.push(P(p1.a+1,p1.b+1));
vis[p1.a+1]=true;
}
if(!vis[p1.a-1]&&p1.a>=1)
{
ans.push(P(p1.a-1,p1.b+1));
vis[p1.a-1]=true;
}
if(!vis[p1.a*2]&&p1.a*2<=100000)
{
ans.push(P(p1.a*2,p1.b+1));
vis[p1.a*2]=true;
}
}
}
}
int main()
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",bfs(a,b));
return 0;
}

【说明】这个代码只在poj上过了，别的oj好像过不了，不知道为啥，显示是wa，但是我感觉过程没问题。还有就是queue可以用数组来简化，因为队列比较慢，可以立两个flag来统计数组模拟queue，这题空间给的很小，不知道可不可行。