【倍增求lca】仓鼠找sugar

 洛谷P3398 仓鼠找sugar

倍增求lca可做qwq

题目大意，只要两条路径有重合之处便输出“Y”;

几点小思路：

1. 如果一对点的LCA的深度比另外一对点中任意一个点的深度都要深，不可能相遇，输出N。
2. 如果一对点的起点or终点与另一对点的起点or终点相同，可能相遇，输出Y。
3. 如果两对点的LCA相同，可能相遇，输出Y。
4. 设r1 = lca(a,b)， r2 = lca(c,d) 且dep[r1] < dep[r2] ， 则若两条路径有重合， lca(r1,c) == r1 || lca(r1,d) == r1; 反之亦然。

代码君qwq

#include<cmath>
#include<cstdio>
#include<iostream>
using namespace std;
const int sz = 100010;
int n, q, log2n;
int num = 0, head[sz], jump[sz][20], dep[sz];
struct node {
    int nxt, to;
}e[sz<<1];
inline int read() {
    int x = 0 , f = 1;char ch = getchar();
    while(ch < '0' || ch > '9'){if(ch == '-') f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9'){x = x * 10 + ch - 48; ch = getchar();}
    return x * f;
}
void add(int from, int to) {
    e[++num].nxt = head[from];
    e[num].to = to;
    head[from] = num;
}
void dfs(int u) {
    for(int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].to;
        if(v != jump[u][0]) {
            jump[v][0] = u;
            dep[v] = dep[u]+1;
            dfs(v);
        }
    }
}
void init() {
    for(int i = 1; i <= log2n; i++) 
        for(int j = 1; j <= n; j++) 
            jump[j][i] = jump[jump[j][i-1]][i-1];
}
int lca(int x, int y) {
    if(dep[x]<dep[y]) swap(x,y);
    int t = dep[x] - dep[y];
    for(int i = 0; (1<<i) <= n; i++) {
        if(t&(1<<i)) x = jump[x][i];
    }
    if(x == y) return x;
    for(int i = log2n; i >= 0; i--) {
        if(jump[x][i] != jump[y][i]) {
            x = jump[x][i];
            y = jump[y][i];
        }
    }
    return jump[x][0];
}
int main() {
    n = read(), q = read();
    log2n = log(n)/log(2)+1;
    for(int i = 1; i < n; i++) {
        int x = read(), y = read();
        add(x, y);
        add(y, x);
    }
    int s = 1;
    dep[s] = 1;
    jump[s][0] = 0;
    dfs(s);
    init();
    for(int i = 1; i <= q; i++) {
        int a = read(), b = read(), c = read(), d = read();
        if(a==c||a==d||b==c||b==d) {
            cout<<"Y"<<endl;
            continue;
        }
        else {
            int r1 = lca(a,b);
            int r2 = lca(c,d);
            if(r1 == r2) {
                cout<<"Y"<<endl;
                continue;
            }
            if(dep[r1]>dep[c] && dep[r1]>dep[d]) {
                cout<<"N"<<endl;
                continue;
            }
            if(dep[r2]>dep[a] && dep[r2]>dep[b]) {
                cout<<"N"<<endl;
                continue;
            }
            if(dep[r1] > dep[r2]) {
                if(lca(r1,c) == r1 || lca(r1,d) == r1) {
                    cout<<"Y"<<endl;
                    continue;
                } 
                else {
                    cout<<"N"<<endl;
                    continue;
                }
            }
            else {
                if(lca(r2,a) == r2 || lca(r2,b) == r2) {
                    cout<<"Y"<<endl;
                    continue;
                } 
                else {
                    cout<<"N"<<endl;
                    continue;
                }
            } 
         }
    }
    return 0;
}

　　　加个快读能快200+ms哦