HDU5266---pog loves szh III （线段树+LCA）

题意：N个点的有向树， Q次询问， 每次询问区间[L, R]内所有点的LCA。

大致做法：线段树每个点保存它的孩子的LCA值， 对于每一次询问只需要 在线段树查询即可。

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3e5+10;
struct Edge{
    int to, next;
}e[MAXN << 1];
int head[MAXN], tot_edge;
void Add_Edge (int x, int y){
    e[tot_edge].to = y;
    e[tot_edge].next = head[x];
    head[x] = tot_edge++;
}
int n, q, MAX_LOG_V;
bool vis[MAXN];
void init (){
    MAX_LOG_V = 20;
    tot_edge = 0;
    memset(head, -1, sizeof (head));
    memset(vis, false, sizeof (vis));
}
int dep[MAXN], pa[25][MAXN];
void DFS (int r, int pre, int d){                    // DFS 或 BFS都可以， G++下BFS
    dep[r] = d;
    pa[0][r] = pre;
    for (int i = head[r]; ~i; i = e[i].next){
        int u = e[i].to;
        if (pre != u){
            DFS(u, r, d+1);
        }
    }
}
void BFS (int r, int pre, int d){
    dep[r] = d;
    pa[0][r] = pre;
    queue <int>Q;
    Q.push(r);
    vis[r] = true;
    while (!Q.empty()){
        int u = Q.front();
        Q.pop();
        for (int i = head[u]; ~i; i = e[i].next){
            int v = e[i].to;
            if (!vis[v]){
                dep[v] = dep[u] + 1;
                pa[0][v] = u;
                Q.push(v);
                vis[v] = true;
            }
        }
    }
}
void pre_solve (){
    BFS(1, -1, 0);
    for (int k = 0; k + 1 < MAX_LOG_V; k++){
        for (int v = 1; v <= n; v++){
            if (pa[k][v] < 0){
                pa[k+1][v] = -1;
            }else{
                pa[k+1][v] = pa[k][pa[k][v]];
            }
        }
    }
}
int LCA (int u, int v){
    if (dep[u] > dep[v]){
        swap(u, v);
    }
    for (int k = 0; k < MAX_LOG_V; k++){
        if ((dep[v] - dep[u]) >> k & 1){
            v = pa[k][v];
        }
    }
    if (u == v){
        return u;
    }
    for (int k = MAX_LOG_V-1; k >= 0; k--){
        if (pa[k][u] != pa[k][v]){
            u = pa[k][u];
            v = pa[k][v];
        }
    }
    return pa[0][u];
}
int seg[MAXN << 2];
void push_up(int pos){
    seg[pos] = LCA(seg[pos<<1], seg[pos<<1|1]);
}
void build (int l, int r, int pos){
    if (l == r){
        seg[pos] = l;
        return ;
    }
    int mid = (l + r) >> 1;
    build(l, mid, pos<<1);
    build(mid+1, r, pos<<1|1);
    push_up(pos);
}
int query (int l, int r, int pos, int ua, int ub){
    if (ua <= l && ub >= r){
        return seg[pos];
    }
    int mid = (l + r) >> 1;
    int t1 = -1, t2 = -1;
    if (ua <= mid){
        t1 = query(l, mid, pos<<1, ua, ub);
    }
    if (ub > mid){
        t2 = query(mid+1, r, pos<<1|1, ua, ub);
    }
    if (t1 == -1 || t2 == -1){
        return max(t1, t2);
    }else{
        return LCA(t1, t2);
    }
}
int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
    #endif
    while (~ scanf ("%d", &n)){
        init();
        for (int i = 0; i < n-1; i++){
            int u, v;
            scanf ("%d%d", &u, &v);
            Add_Edge(u, v);
            Add_Edge(v, u);
        }
        pre_solve();
        build(1, n, 1);
        scanf ("%d", &q);
        for (int i = 0; i < q; i++){
            int ua, ub;
            scanf ("%d%d", &ua, &ub);
            printf("%d
", query(1, n, 1, ua, ub));
        }
    }
    return 0;
}