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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 52537 | Accepted: 16471 |
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
Source
ac代码
#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
#define INF 0xfffffff
#define min(a,b) (a>b?b:a)
using namespace std;
struct s
{
int pos,step;
}a,temp;
int n,k,ans,vis[100100];
void bfs()
{
memset(vis,0,sizeof(vis));
a.pos=n;
a.step=0;
queue<struct s>q;
q.push(a);
vis[a.pos]=1;
while(!q.empty())
{
a=q.front();
q.pop();
for(int i=0;i<3;i++)
{
if(i==0)
temp.pos=a.pos+1;
if(i==1)
temp.pos=a.pos-1;
if(i==2)
temp.pos=a.pos*2;
temp.step=a.step+1;
if(temp.pos==k)
{
ans=min(ans,temp.step);
continue;
}
if(temp.pos<0||temp.pos>100001)
continue;
if(!vis[temp.pos])
{
vis[temp.pos]=1;
q.push(temp);
}
}
}
}
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
ans=INF;
if(n==k)
{
printf("0\n");
continue;
}
if(n>k)
{
printf("%d\n",n-k);
continue;
}
bfs();
printf("%d\n",ans);
}
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原文地址：http://blog.csdn.net/yu_ch_sh/article/details/44520057
