poj3278-Catch That Cow 【bfs】

http://poj.org/problem?id=3278

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 52463		Accepted: 16451

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.

 

 

 

题解：最短路，第一想法是bfs，简单易操作。这里需要说明一下就是，这道题目的边界不需要扩充，理由嘛，看下面0ms代码的注释吧。再次强调，STL少用，这里再次验证了其效率的低下。

代码一：【127ms】bfs暴搜+STL里的queue：

 #include <fstream>
 #include <iostream>
 #include <algorithm>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <cstdlib>
 #include <queue>

 using namespace std;

 #define PI acos(-1.0)
 #define EPS 1e-10
 #define lll __int64
 #define ll long long
 #define INF 0x7fffffff

 queue<pair<int,int> > qu;
 bool b[];

 inline bool Check(int x);
 int Bfs(int r1,int r2);

 int main(){
     //freopen("D:\\input.in","r",stdin);
     //freopen("D:\\output.out","w",stdout);
     int r1,r2;
     scanf("%d %d",&r1,&r2);
     printf("%d\n",Bfs(r1,r2));
     return ;
 }
 inline bool Check(int x){
     return x>=&&x<=&&b[x]==;
 }
 int Bfs(int r1,int r2){
     qu.push(make_pair(r1,));
     b[r1]=;
     while(!qu.empty()){
         pair<int,int> tmp=qu.front();
         qu.pop();
         if(tmp.first==r2){
             return tmp.second;
         }
         int x=tmp.first+;
         if(Check(x)){
             b[x]=;
             qu.push(make_pair(x,tmp.second+));
         }
         x=tmp.first-;
         if(Check(x)){
             b[x]=;
             qu.push(make_pair(x,tmp.second+));
         }
         x=tmp.first*;
         if(Check(x)){
             b[x]=;
             qu.push(make_pair(x,tmp.second+));
         }
     }
 }

我自己底下实验了下，把STL去掉，自己实现简易的队列，再进行暴搜，时间消耗为16ms。

代码二：【0ms】bfs剪枝(这里没有用STL)：         剪枝内容：当自己的位置大于k的时候，自然不用考虑+1和*2的情况了。

 #include <fstream>
 #include <iostream>
 #include <cstdio>

 using namespace std;

 const int N=;
 int n,k,dis[N],queue[N];//dis:记录走到某个点的步数。 queue:因为每次队列采用去重加入，所以N是足够用了。
 bool b[N];//N是足够的，不需要扩展2N或者3N。打比说x先乘再减，与先减再乘，后者会更快，故不会超过N，负数是更不可能的，因为没有除法。

 int bfs();

 int main()
 {
     //freopen("D:\\input.in","r",stdin);
     //freopen("D:\\output.out","w",stdout);
     scanf("%d%d",&n,&k);
     printf("%d",bfs());
     return ;
 }
 int bfs(){
     int l,r,x;
     l=r=;
     dis[n]=;
     queue[]=n;
     while(){//队列不可能为空
         if(queue[l]==k) return dis[k];
         if(x=queue[l]-,x>=&&!b[x]){//注意边界
             b[x]=;
             queue[++r]=x;
             dis[x]=dis[queue[l]]+;
         }
         if(x=queue[l]+,queue[l]<=k&&!b[x]){//这里的queue[l]<=k为剪枝
             b[x]=;
             queue[++r]=x;
             dis[x]=dis[queue[l]]+;
         }
         if(x=queue[l]<<,queue[l]<=k&&x<N&&!b[x]){//这里的queue[l]<=k为剪枝；注意边界
             b[x]=;
             queue[++r]=x;
             dis[x]=dis[queue[l]]+;
         }
         l++;
     }
 }