| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 51447 | Accepted: 16134 |
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
Hint
题意：
给出起始位置n，有三种操作x+1,x-1,2*x,求最少的操作次数，使x==k。
题解：
广搜，裸的广搜会MLE，要加剪枝，还要用数组标记。一直MLE，就调整为每次出现一个新位置就判断是否结束，导致后来忘记考虑n==k的情况%>_<%
剪枝1.x>k,2*x不用走，只会越走越远。
剪枝2.x>k,x+1不用走，使x变小的只有x-1，x+1再x-1只会增加步数。
剪枝3.x-1>=0,x为负数时，只会增加步数或越走越远。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
int n,k;
int ans;
const int maxn=200000+100;
struct node
{
int x;
int cur;
} p;
bool vis[maxn];
queue<node> q;
void BFS()
{
node p;
p.x=n;
p.cur=0;
vis[n]=true;
while(!q.empty())
q.pop();
q.push(p);
while(!q.empty())
{
node now;
node tmp=q.front();
q.pop();
if(tmp.x==k)//n==k时
{
ans=tmp.cur;
return;
}
if(tmp.x<k&&!vis[tmp.x*2])//剪枝
{
now.x=tmp.x*2;
now.cur=tmp.cur+1;
vis[now.x]=true;
if(now.x==k)
{
ans=now.cur;
return;
}
q.push(now);
}
if(tmp.x<k&&!vis[tmp.x+1])//剪枝
{
now.x=tmp.x+1;
now.cur=tmp.cur+1;
vis[now.x]=true;
if(now.x==k)
{
ans=now.cur;
return;
}
q.push(now);
}
if(tmp.x-1>=0&&!vis[tmp.x-1])//剪枝
{
now.x=tmp.x-1;
now.cur=tmp.cur+1;
vis[now.x]=true;
if(now.x==k)
{
ans=now.cur;
return;
}
q.push(now);
}
}
}
int main()
{
while(~scanf("%d%d",&n,&k))
{
memset(vis,false,sizeof(vis));
BFS();
printf("%d\n",ans);
}
return 0;
}
原文地址：http://blog.csdn.net/caduca/article/details/43668185
