POJ 3278 Catch That Cow <BFS>

题目：传送门
分析：超简单的bfs，稍微剪枝一下就好了，比如X-1要判断>=0，X+1要判断<=k，X*2要判断小于K*2。最后注意的就是N==K这种情况。
代码：

#include <iostream>#include <algorithm>#include <queue>#include <cstring>using namespace std;const int MAXN=2e5+10;int n,k;struct Node{    int x,time;    Node(int _x,int _time):x(_x),time(_time){}};queue<Node> q;bool vis[MAXN];void bfs(){    if(n==k){        cout<<0<<endl;        return ;    }    while(!q.empty()) q.pop();    memset(vis,false,sizeof vis);    q.push(Node(n,0));    vis[n]=true;    while(!q.empty()){        Node cur=q.front();        q.pop();        int xx=cur.x+1;        if(xx==k){            cout<<cur.time+1<<endl;            return ;        }        if(xx<k&&!vis[xx]){            vis[xx]=true;            q.push(Node(xx,cur.time+1));        }        xx=cur.x-1;        if(xx==k){            cout<<cur.time+1<<endl;            return ;        }        if(xx>=0&&!vis[xx]){            vis[xx]=true;            q.push(Node(xx,cur.time+1));        }        xx=cur.x*2;        if(xx==k){            cout<<cur.time+1<<endl;            return ;        }        if(xx<k*2&&!vis[xx]){            vis[xx]=true;            q.push(Node(xx,cur.time+1));        }    }}int main(){    cin>>n>>k;    bfs();    return 0;}