P3398 仓鼠找sugar (一道LCA的裸题)

https://www.luogu.org/problemnew/show/P3398

题意简单概括一下就是求树上两条路径是否相交;

有这样一个性质:

if相交,则必有lca(a,b) 在路径c <-> d 上or lca(c,d) 在路径a <-> b 上;

接下来就是这样一个问题:

怎样判断一个点(x)是否在一条路径(a,b)上

如果满足以下两个条件,则在:

1.dep[x] >= lca(a,b);

2.lca(x,a) == x || lca(x,b) == x

#include <bits/stdc++.h>
#define read read()
#define up(i,l,r) for(register int i = (l);i <= (r);++i)
#define down(i,l,r) for(register int i = (l);i >= (r);i--)
#define traversal_vedge(i) for(register int i = head[u]; i ;i = e[i].nxt)
#define ll long long
using namespace std;
int read
{
    int x = , f = ; char ch = getchar();
    while(ch <  || ch > ) {if(ch == '-')f = -; ch = getchar();}
    while(ch >= && ch <=) {x =  * x + ch - ;ch = getchar();}
    return x * f;
}
//-----------------------------------------------------------------
const int N = ;
int n,q;

struct edge{
    int v,nxt;
}e[N<<];int tot,head[N];

void buildtree(int u,int v) {e[++tot] = (edge){v,head[u]}; head[u] = tot;}
//-----------------------------------------------------------------

int dep[N],size[N],top[N],fa[N];
void initdfs(int u){
    dep[u] = dep[fa[u]] + ;
    size[u] = ; top[u] = u;
    int hson_id = ,hson_size = ;
    traversal_vedge(i)
    {
        int v = e[i].v;
        if(fa[u] == v) continue;
        fa[v] = u;
        initdfs(v);
        size[u] += size[v];
        if(size[v] > hson_size) hson_id = v,hson_size = size[v];
    }
    if(hson_id) top[hson_id] = u;
}
int find(int u){if(top[u] == u) return u;top[u] = find(top[u]); return top[u];}
int lca(int x,int y){
    if(find(x) != find(y)){
        if(dep[top[x]] > dep[top[y]]) return lca(fa[top[x]],y);
        else return lca(x,fa[top[y]]);
    }
    return dep[x] > dep[y] ? y : x;
}
//-----------------------------------------------------------------
bool ok;
void query(int x,int a,int b){
    if(lca(a,x) == x || lca(b,x) == x) ok = ;
}

void work(){
    initdfs();
    while(q--)
    {
        ok = ;
        int a = read,b = read,c = read,d = read;
        int x = lca(a,b);int y = lca(c,d);//debug y = lca(a,b)
        if(dep[x] >= dep[y]) query(x,c,d);
        if(!ok) if(dep[y] >= dep[x]) query(y,a,b);
        if(ok == ) printf("Y\n");
         else printf("N\n");
     }
} 

void readdata(){
    n = read; q = read;
    up(i,,n-)
    {
        int u = read,v = read;
        buildtree(u,v);
        buildtree(v,u);
    }
}

int main()
{
    readdata();
    work();
    return ;
}