1-Trees and Queries

E. 1-Trees and Queries

该题需要用 lca 进行优化。

在树中求两点间的距离的方法为：dis=dep[l]+dep[r]-2*dep[lca(l,r)]

// Created by CAD on 2020/2/16.
#include <bits/stdc++.h>
using namespace std;

const int maxn=1e5+5;
int n;
//-----------------前向星--------------------
int cnt=0,head[maxn<<1];
struct edge{
    int to,w,next;
}e[maxn<<1];
void add(int u,int v,int w){
    e[++cnt]=edge{v,w,head[u]};
    head[u]=cnt;
}
//--------------------树链剖分----------------------
int dep[maxn],id[maxn],siz[maxn],son[maxn],f[maxn],top[maxn],w[maxn];
int tot;
void dfs1(int x,int fa,int deep){
    dep[x]=deep,f[x]=fa,siz[x]=1;
    int maxson=-1;
    for(int i=head[x];i;i=e[i].next){
        int v=e[i].to;
        if(v==fa) continue;
        w[v]=e[i].w;
        dfs1(v,x,deep+1);
        siz[x]+=siz[v];
        if(siz[v]>maxson) maxson=siz[v],son[x]=v;
    }
}
void dfs2(int x,int from){
    id[x]=++tot;
    top[x]=from;
    if(son[x]) dfs2(son[x],from);
    for(int i=head[x];i;i=e[i].next){
        int v=e[i].to;
        if(v==son[x]||v==f[x]) continue;
        dfs2(v,v);
    }
}
int qrange(int l,int r){
    int ans=dep[l]+dep[r];
    while(top[l]!=top[r]){
        if(dep[top[l]]<dep[top[r]]) swap(l,r);
        l=f[top[l]];
    }
    if(dep[l]>dep[r]) swap(l,r);
    ans-=(2*dep[l]);
    return ans;
}
int main()
{
    scanf("%d", &n);
    for (int i=1; i<=n-1; ++i)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        add(u, v, 1), add(v, u, 1);
    }
    w[1]=1;
    dfs1(1, 0, 1), dfs2(1, 1);
    int q;  scanf("%d", &q);
    while (q--)
    {
        int x, y, a, b, k;
        scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
        int ab=qrange(a, b);
        bool ok=0;
        if (ab<=k && ab%2==k%2) ok=1;
        else
        {
            int ax=qrange(a, x), ay=qrange(a, y), xb=qrange(x, b), yb=qrange(y, b);
            int r1=ax+yb+1, r2=ay+xb+1;
            if (r1<=k && r1%2==k%2) ok=1;
            if (r2<=k && r2%2==k%2) ok=1;
        }
        puts(ok ? "YES" : "NO");
    }
    return 0;
}