| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 62146 | Accepted: 19460 |
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
Output
Sample Input
5 17
Sample Output
4
题目大意：就是给你两个数 n 和 k 让你通过以下三种方法使得 n -> k;
1) n + 1;
2) n - 1；
3）n * 2;
看怎么样通过上述三种方法使得 n 变成 k 而且需要的步骤最少；
样例解释：
5 17
5 -> 10(1) -> 9(2) -> 18(3) -> 17(4)
所以只需要 4 步
解题思路：这是广搜的类型，所以需要bfs就行了，具体注意的问题会在代码里给出：
/*
Date : 2015-09-08 晚上
Author : ITAK
Motto :
今日的我要超越昨日的我，明日的我要胜过今日的我；
以创作出更好的代码为目标，不断地超越自己。
*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 100001;
bool vis[maxn];
int n,k;
int fac[maxn];///记录结果
int dis[maxn];//
void bfs()
{
int front, rear, dx, dy;
rear = front = 0;
dis[0] = n, fac[n] = 0;
vis[n] = 1;
while(front <= rear)
{
dx = dis[front];
dy = dx - 1;
if(dy>=0 && !vis[dy])///剪枝 -> dy>=0
{
rear++;
dis[rear] = dy;
fac[dy] = fac[dx] + 1;
vis[dy] = 1;
}
dy = dx + 1;
if(dy<maxn && !vis[dy])///剪枝 -> dy < maxn
{
rear++;
dis[rear] = dy;
fac[dy] = fac[dx] + 1;
vis[dy] = 1;
}
dy = dx*2;
if(dy<maxn && !vis[dy])///剪枝 -> dy < maxn
{
rear++;
dis[rear] = dy;
fac[dy] = fac[dx] + 1;
vis[dy] = 1;
}
if(vis[k])///结束
break;
front++;
}
}
int main()
{
while(~scanf("%d%d",&n,&k))
{
memset(vis, 0, sizeof(vis));
bfs();
cout<<fac[k]<<endl;
}
return 0;
}
