POJ 3278 Catch That Cow
Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 64397 | Accepted: 20226 |
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
= =|| 训练队同队的队员AC之后一直在向我炫耀 估计也是做不成队友了（手动再见）
初学STL写的不太好，刚开始还忘记了n>k的情况
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
using namespace std;
bool vis[200001];
typedef struct
{
int data,step;
}position;
position father,current;
queue<position> bus;
int vplus,vreduce,vadd;
int main(void)
{
int n,k;
scanf("%d %d",&n,&k);
if(n==k)
{
printf("0\n");
return 0;
}
if(n>k)
{
printf("%d\n",n-k);
return 0;
}
current.step=0;
current.data=n;
bus.push(current);
while(!bus.empty())
{
father=bus.front();
bus.pop();
if(father.data==k)
{
printf("%d\n",father.step);
return 0;
}
vis[father.data]=true;
vplus=father.data+father.data;
vreduce=father.data-1;
vadd=father.data+1;
current.step=father.step+1;
if(vplus<200001&&!vis[vplus])
{
current.data=vplus;
bus.push(current);
}
if(vadd<200001&&!vis[vadd])
{
current.data=vadd;
bus.push(current);
}
if(vreduce>-1&&!vis[vreduce])
{
current.data=vreduce;
bus.push(current);
}
}
return (0);
}
下面是在其他地方找到的一份代码，也AC了不过时间好短w
#include<iostream>
#include<queue>
using namespace std;
#define M 100001
#define INF 10000000
struct POINT
{
int pos;
int step;
}now,next;
queue<POINT>Q;
bool visited[M];
int n,k;
int bfs()
{
while(!Q.empty())
Q.pop();
now.pos=n;
now.step=0;
visited[now.pos]=true;
Q.push(now);
while(!Q.empty())
{
now=Q.front();
Q.pop();
next=now;
for(int i=0;i<3;i++)
{
if(i==0)
next.pos=now.pos+1;
if(i==1)
next.pos=now.pos-1;
if(i==2)
next.pos=now.pos*2;
next.step=now.step+1;
if(next.pos==k)
return next.step;
if(next.pos<0||next.pos>M)
continue;
if(!visited[next.pos])
{
visited[next.pos]=true;
Q.push(next);
}
}
}
return INF;
}
int main()
{
while(cin>>n>>k)
{
memset(visited,false,sizeof(visited));
if(n<k)
cout<<bfs()<<endl;
if(n==k)
cout<<0<<endl;
if(n>k)
cout<<n-k<<endl;
}
}
>_< 继续学习www