[JZOJ 5852] 相交

Posted 2021-01-04 akoasm

题意：求树上两条路径有无祖先。
思路：
瞎搞(LCA)啊。。。
可惜我(LCA)打错了，我居然调了半小时...qwq

#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
#define travel(i,x) for(int i = head[x];i;i=e[i].nxt)
inline int read () {
    int q=0,f=1;char ch = getchar();
    while(!isdigit(ch)){
        if(ch==‘-‘)f=-1;ch=getchar();
    }
    while(isdigit(ch)){
        q=q*10+ch-‘0‘;ch=getchar();
    }
    return q*f;
} 
struct edge {
    int to;
    int nxt;
}e[maxn << 1];
int cnt;
int head[maxn << 1];
inline void add(int u,int v) {
    e[++cnt].to = v;
    e[cnt].nxt = head[u];
    head[u] = cnt;
}
int f[maxn][30];
int dep[maxn << 1];
inline void dfs(int x,int fa) {
    dep[x] = dep[fa] + 1;
    travel(i,x) {
        int y = e[i].to;
        if(y != fa) {
            f[y][0] = x;
            for(int j = 1;j <= 20; ++j) {
                f[y][j] = f[f[y][j - 1]][j - 1];
            }
            dfs(y,x);
        }
    }
}
inline int lca(int x,int y) {
    if(dep[x] < dep[y]) swap(x,y);
    for(int i = 20;i >= 0; --i) {
        if(dep[f[x][i]] >= dep[y]) x = f[x][i];
    }
    if(x == y) return x;
    for(int i = 20; i >= 0; --i) {
        if(f[x][i] != f[y][i]) {
            x = f[x][i];
            y = f[y][i];
        }
    }
    return f[x][0];
}
int n,m;
int main () {
    freopen("inter.in","r",stdin);
    freopen("inter.out","w",stdout);
    n = read();
    for(int i = 1;i < n; ++i) {
        int x = read(),y = read();
        add(x,y);add(y,x);
    }
    dfs(1,0);
    m = read();
    while(m--) {    
        int x = read(),y = read();
        int LCA = lca(x,y);
        int l = read(),r = read();
        int _LCA = lca(l,r);
        //cout<<"LCA:"<<LCA<<‘ ‘<<_LCA<<endl;
        int Grand_LCA = lca(LCA,_LCA);
        if(LCA == _LCA) {
            puts("YES");
        }
        else if(Grand_LCA == _LCA && (lca(l,LCA) == LCA || lca(r,LCA) == LCA)) {
            puts("YES");
        }
        else if(Grand_LCA == LCA && (lca(x,_LCA) == _LCA || lca(y,_LCA) == _LCA)) {
            puts("YES");
        }
        else puts("NO");
    }
    return 0;
}