Luogu P3398 仓鼠找sugar 倍增LCA

题目描述
题目

差不多是裸的LCA吧
只要处理好两条相交路径之间的关系就好了
如果两条路径相交，那么一条路径的LCA一定在另一条路径上
那么需要满足

dep[lca1] >= dep[lca2]
lca(lca1,s) == lca1 || lca(lca1,t) == lca1

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=100010;
int n,q,now,head[N],f[N][20],dep[N];
struct E{int to,next;}e[N<<1];
void build(int u,int v){e[++now].to=v;e[now].next=head[u];head[u]=now;}

void dfs1(int u,int fa)
{
    f[u][0]=fa;
    for(int i=1;f[u][i-1];i++) f[u][i]=f[f[u][i-1]][i-1];
    for(int i=head[u];i;i=e[i].next)
    {
        int v=e[i].to;
        if(v == fa) continue;
        dep[v]=dep[u]+1;
        dfs1(v,u);
    }
}

int lca(int u,int v)
{
    if(dep[u] > dep[v]) swap(u,v);
    int d=dep[v]-dep[u];
    for(int i=0;(1<<i) <= d;i++)
        if((1<<i)&d) v=f[v][i];
    if(u == v) return u;
    for(int i=19;i >= 0;i--)
        if(f[u][i] != f[v][i])
            u=f[u][i],v=f[v][i];
    return f[u][0];
}

int read(){
    int out=0;char c=getchar();while(c > '9' || c < '0') c=getchar();
    while(c >= '0' && c <= '9') {out=(out<<1)+(out<<3)+c-'0';c=getchar();}return out;
}

void init()
{
    n=read(),q=read();
    for(int i=1;i<n;i++) {int u=read(),v=read();build(u,v);build(v,u);}
    dfs1(1,0);
}

void solve()
{
    for(int i=1;i<=q;i++)
    {
        int u1=read(),v1=read(),u2=read(),v2=read();
        int lca1=lca(u1,v1),lca2=lca(u2,v2);
        if(lca1 == lca2){printf("Y\n");continue;}
        if(dep[lca2] > dep[u1] && dep[lca2] >dep[v1]) {printf("N\n");continue;}
        if(dep[lca1] > dep[u2] && dep[lca1] >dep[v2]) {printf("N\n");continue;}
        if(dep[lca1] >= dep[lca2])
        {
            if(lca1 == lca(lca1,u2)) {printf("Y\n");continue;}
            if(lca1 == lca(lca1,v2)) {printf("Y\n");continue;}
            printf("N\n");continue;
        }
        if(dep[lca2] >= dep[lca1])
        {
            if(lca2 == lca(lca2,u1)) {printf("Y\n");continue;}
            if(lca2 == lca(lca2,v1)) {printf("Y\n");continue;}
            printf("N\n");continue;
        }
        printf("N\n");
    }
}

int main()
{
    init();
    solve();
    return 0;
}