POJ - 3278 Catch That Cow

题目：

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

目前只尝试了BFS，不会超时。

#include<iostream>
#include<queue>
using namespace std;

int line[200000]={0};//number line ，这里选择最大长度的2倍
int N,K;
queue<int> q; //BFS队列

int main(){
    cin >>N>>K;
    if(N==K){ //这里是一个坑！要单独判断N==K的情况
        cout <<0<<endl;
        system("pause");
        return 0;
    }
    //BFS
    q.push(N);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        int choice[3];
        choice[0]=u-1; //三种可能的移动目标位置
        choice[1]=u+1;
        choice[2]=2*u;
        for(int i=0;i<3;i++){
            if(choice[i]>=0&&choice[i]<200000&&line[choice[i]]==0){
                line[choice[i]]=line[u]+1;
                q.push(choice[i]);
                if(choice[i]==K){ //如果到达K点就结束
                    cout <<line[choice[i]]<<endl;
                    break;
                }
            }
        }
    }
    system("pause");
    return 0;
}

 

需要注意的一点就是边界情况要记得去考虑一下有没有包含进来，这里N==K这个情况没有考虑到的话就会WA