AtCoder Beginner Contest 148 F.Playing Tag on Tree

AtCoder Beginner Contest 148 F.Playing Tag on Tree

题目链接

Problem Statement

（略）

Output

Print the number of moves Aoki will perform before the end of the game.

Sample Input 1

5 4 1
1 2
2 3
3 4
3 5

Sample Output 1

2

Sample Input 2

5 4 5
1 2
1 3
1 4
1 5

Sample Output 2

1

Sample Input 3

2 1 2
1 2

Sample Output 3

0

Sample Input 4

9 6 1
1 2
2 3
3 4
4 5
5 6
4 7
7 8
8 9

Sample Output 4

5

题意比较简单，但是比较难下手，看了题解才知道要找的结点满足离Aoki最远且Takahashi到达该结点比Aoki早即可，就跑了两遍BFS，最后遍历更新一遍即可，AC代码如下：

#include 
#define maxn 100005
using namespace std;
typedef long long ll;

vector<int>G[maxn];
int step1[maxn],step2[maxn];
int vis1[maxn],vis2[maxn];
int n,s1,s2,u,v;

void bfs(int k){
    if(k==s1){
        queue<int>q;
        q.push(k);
        step1[k]=0;
        vis1[k]=1;
        while(!q.empty()){
          int a=q.front();
          q.pop();
          for(int b:G[a]){
             if(!vis1[b]) {
                step1[b]=step1[a]+1;
                vis1[b]=1;
                q.push(b);
          }
        }
      }
    }
    else if(k==s2){
        queue<int>q;
        q.push(k);
        step2[k]=0;
        vis2[k]=1;
        while(!q.empty()){
          int a=q.front();
          q.pop();
          for(int b:G[a]){
            if(!vis2[b]) {
                step2[b]=step2[a]+1;
                vis2[b]=1;
                q.push(b);
            }
         }
      }
   }
}

int main(){
    memset(vis1,0,sizeof(vis1));
    memset(vis2,0,sizeof(vis2));
    memset(step1,0,sizeof(step1));
    memset(step2,0,sizeof(step2));
    cin>>n>>s1>>s2;
    for(int i=0;i<n-1;i++){
        scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    bfs(s1);
    bfs(s2);
    int ans=0;
    for(int i=1;i<=n;i++){
        if(step1[i]<step2[i]) ans=max(ans,step2[i]);
    }
    cout<<ans-1;
  return 0;
}