Codeforces 1304E. 1-Trees and Queries 代码（LCA 树上两点距离判奇偶）

https://codeforces.com/contest/1304/problem/E

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxbit = ;
 const int maxn = 1e5+;
 vector<int> G[maxn];
 int depth[maxn];
 int fa[maxn][maxbit];
 int Log[maxn];
 int N;
 void pre(){
     Log[] = -;
     Log[] = ,Log[] = ;
     for(int i = ;i<maxn;i++) Log[i] = Log[i/] + ;
 }
 void dfs(int cur,int father){//dfs预处理
     depth[cur] = depth[father] + ;//当前结点的深度为父亲结点+1
     fa[cur][] = father;//更新当前结点的父亲结点
     for(int j = ;(<<j)<=N;j++){//倍增更新当前结点的祖先
         fa[cur][j] = fa[fa[cur][j-]][j-];
     }
     for(int i = ;i<G[cur].size() ;i++){
         if(G[cur][i] != father) {//dfs遍历
             dfs(G[cur][i],cur);
         }
     }
 }
 int LCA(int u,int v){
     if(depth[u]<depth[v]) swap(u,v);
     int dist = depth[u] - depth[v];//深度差
     while(depth[u]!=depth[v]){//把较深的结点u倍增到与v高度相等
         u = fa[u][Log[depth[u]-depth[v]]];
     }
     if(u == v) return u;//如果u倍增到v，说明v是u的LCA
     for(int i = Log[depth[u]];i>=;i--){//否则两者同时向上倍增
         if(fa[u][i]!=fa[v][i]){//如果向上倍增的祖先不同，说明是可以继续倍增
             u = fa[u][i];//替换两个结点
             v = fa[v][i];
         }
     }
     return fa[u][];//最终结果为u v向上一层就是LCA
 }
 int main()
 {
     pre();
     cin>>N;
     for(int i = ;i<N;i++){
         int u,v;
         cin>>u>>v;
         G[u].push_back(v),G[v].push_back(u);
     }
     dfs(,);
     int q;cin>>q;
     while(q--){
         int x,y,a,b,k;
         cin>>x>>y>>a>>b>>k;
         int t1 = LCA(a,b),t2 = LCA(a,x),t3 = LCA(a,y),t4 = LCA(b,x),t5  = LCA(b,y);
         int dis = abs(depth[t1]-depth[a])+abs(depth[t1]-depth[b]);
         if(dis% == k% && dis <=k ){
 //            cout<<"De1"<<endl;
             cout<<"YES"<<endl;
             continue;
         }
         int dis1 = abs(depth[t2]-depth[a])+abs(depth[t2]-depth[x]);
         int dis2 = abs(depth[t5]-depth[b])+abs(depth[t5]-depth[y]);
         if( (dis1+dis2+)% == k%  && dis1+dis2+ <= k){
 //            cout<<"de2"<<endl;
             cout<<"YES"<<endl;
             continue;
         }
         dis1 = abs(depth[t3]-depth[a])+abs(depth[t3]-depth[y]);
         dis2 = abs(depth[t4]-depth[b])+abs(depth[t4]-depth[x]);
         if((dis1+dis2+)% == k% && dis1+dis2+ <= k){
 //            cout<<"de3"<<endl;
             cout<<"YES"<<endl;
             continue;
         }
         cout<<"NO"<<endl;
     }
     return ;
 }