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[100002];

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
",Bfs(r1,r2));
    return 0;
}
inline bool Check(int x){
    return x>=0&&x<=100000&&b[x]==0;
}
int Bfs(int r1,int r2){
    qu.push(make_pair(r1,0));
    b[r1]=1;
    while(!qu.empty()){
        pair<int,int> tmp=qu.front();
        qu.pop();
        if(tmp.first==r2){
            return tmp.second;
        }
        int x=tmp.first+1;
        if(Check(x)){
            b[x]=1;
            qu.push(make_pair(x,tmp.second+1));
        }
        x=tmp.first-1;
        if(Check(x)){
            b[x]=1;
            qu.push(make_pair(x,tmp.second+1));
        }
        x=tmp.first*2;
        if(Check(x)){
            b[x]=1;
            qu.push(make_pair(x,tmp.second+1));
        }
    }
}

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=100005;
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 0;
}
int bfs(){
    int l,r,x;
    l=r=0;
    dis[n]=0;
    queue[0]=n;
    while(1){//队列不可能为空
        if(queue[l]==k) return dis[k];
        if(x=queue[l]-1,x>=0&&!b[x]){//注意边界
            b[x]=1;
            queue[++r]=x;
            dis[x]=dis[queue[l]]+1;
        }
        if(x=queue[l]+1,queue[l]<=k&&!b[x]){//这里的queue[l]<=k为剪枝
            b[x]=1;
            queue[++r]=x;
            dis[x]=dis[queue[l]]+1;
        }
        if(x=queue[l]<<1,queue[l]<=k&&x<N&&!b[x]){//这里的queue[l]<=k为剪枝；注意边界
            b[x]=1;
            queue[++r]=x;
            dis[x]=dis[queue[l]]+1;
        }
        l++;
    }
}