POJ3278 Catch That Cow(BFS入門)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 43276		Accepted: 13472

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

解題報告：

入門級廣搜，寫了兩種隊列，一個使用C++ STL裏面的queue，另一種是手寫的隊列，好像都差不多，練練手而已啦！。。

代碼一（STL queue）：

#include<stdio.h>
#include<queue>
#include<algorithm>
#include<string.h>
#define MAXN 100010
using namespace std;
int d[MAXN],vis[MAXN];
int main()
{
  int n,k;
  scanf("%d%d",&n,&k);
  for(int i=0;i<=k;i++)vis[i]=0;
  queue<int>q;
  int v=n;
  d[v]=0;vis[v]=1;
  q.push(v);
  while(!q.empty())
  {
    if(v==k)break;
    v=q.front();
    q.pop();
    if(v+1<MAXN&&!vis[v+1])
    {
      d[v+1]=d[v]+1;
      vis[v+1]=1;
      q.push(v+1);
    }
    if(v-1>=0&&!vis[v-1])
    {
      d[v-1]=d[v]+1;
      vis[v-1]=1;
      q.push(v-1);
    }
    if(v*2<MAXN&&!vis[v*2])
    {
      d[v*2]=d[v]+1;
      vis[v*2]=1;
      q.push(v*2);
    }
    if(v+1==k)break;
    if(v-1==k)break;
    if(v*2==k)break;
  }
  printf("%d\n",d[k]);
  return 0;
}

代碼二（手寫的隊列）：

#include<stdio.h>
#define MAXN 100010
using namespace std;
int main()
{
  int d[MAXN],vis[MAXN],n,k;
  while(scanf("%d%d",&n,&k)!=EOF)
  {
  int que[2*MAXN],front,rear;
  for(int i=0;i<=k;i++)vis[i]=0;
  int v=n;
  front=0;rear=1;
  que[front]=v;
  d[v]=0;
  while(front<rear)
  {
    if(v==k)break;
    v=que[front++];
    if(v+1<MAXN&&!vis[v+1])
    {
      vis[v+1]=1;
      d[v+1]=d[v]+1;
      que[rear++]=v+1;
      if(v+1==k)break;
    }
    if(v-1>=0&&!vis[v-1])
    {
      vis[v-1]=1;
      d[v-1]=d[v]+1;
      que[rear++]=v-1;
      if(v-1==k)break;
    }
    if(v*2<MAXN&&!vis[v*2])
    {
      vis[v*2]=1;
      d[v*2]=d[v]+1;
      que[rear++]=v*2;
      if(v*2==k)break;
    }
  }
  printf("%d\n",d[k]);
  }
  return 0;
}