双向广搜 codevs 3060 抓住那头奶牛

 1 /*借助这道题目再练习一下双向广搜*/
 2 #include<iostream>
 3 using namespace std;
 4 #define N 200100
 5 #include<cstdio>
 6 int visit[N];
 7 int dis[N]={0};
 8 #include<queue>
 9 struct poi{
10     int x,tim,flag;
11 };
12 int n,k;
13 int bfs()
14 {
15     queue<poi>que;
16     visit[n]=1;visit[k]=2;
17     dis[n]=0;dis[k]=0;
18     que.push(poi{n,0,1});
19     que.push(poi{k,0,2});
20     while(!que.empty())
21     {
22         poi now=que.front();
23         que.pop();
24         if(now.x-1>=0&&!visit[now.x-1])
25         {
26             visit[now.x-1]=now.flag;
27             dis[now.x-1]=now.tim+1;
28             que.push(poi{now.x-1,now.tim+1,now.flag});
29         }
30         else if(visit[now.x-1]&&now.flag!=visit[now.x-1])
31              {
32                  return dis[now.x-1]+1+now.tim;
33              }
34         if(now.x+1<=(N/2)&&!visit[now.x+1])
35         {
36             visit[now.x+1]=now.flag;
37             dis[now.x+1]=now.tim+1;
38             que.push(poi{now.x+1,now.tim+1,now.flag});
39         }
40         else if(visit[now.x+1]&&now.flag!=visit[now.x+1])
41              {
42                  return dis[now.x+1]+1+now.tim;
43              }
44         if(now.x*2<=(N/2)&&!visit[now.x*2])
45         {
46             visit[now.x*2]=now.flag;
47             dis[now.x*2]=now.tim+1;
48             que.push(poi{now.x*2,now.tim+1,now.flag});
49         }
50         else if(visit[now.x*2]&&now.flag!=visit[now.x*2])
51              {
52                  return dis[now.x*2]+1+now.tim;
53              }
54     }
55 }
56 int main()
57 {
58     scanf("%d%d",&n,&k);
59     cout<<bfs()<<endl;
60     return 0;
61 }