hdu 5266 pog loves szh III LCA+RMQ

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题意：

给你一棵树，然后询问l~r节点的最近公共祖先（LCA）。

思路：

用RMQ维护一段区间的LCA，然后询问时，将两个区间的LCA再求一次LCA即可。

code：

#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdlib>#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int N = 3e5+5;typedef long long LL;int n, q;int head[N], cnt;struct Edge {    int v, next;}edge[N<<1];int par[N][30];int dep[N];int d[N];int rmq[N][30];void init() {    memset(head, -1, sizeof(head));    cnt = 0;}Edge t;void addEdge(int u, int v) {    t.v = v, t.next = head[u];    edge[cnt] = t;    head[u] = cnt++;}void dfs(int u, int p, int d) {    dep[u] = d; par[u][0] = p;    for(int i = head[u];i != -1; i = edge[i].next) {        int v = edge[i].v;        if(v == p) continue;        dfs(v, u, d+1);    }}    int lca(int u, int v) {    if(dep[u] > dep[v]) swap(u, v);    for(int i = 0;i < 20; i++) {        if(((dep[v]-dep[u])>>i) & 1)            v = par[v][i];    }    if(u == v) return v;    for(int k = 20;k >= 0; k--)        if(par[v][k] != par[u][k]) {            u = par[u][k];            v = par[v][k];        }    return par[u][0];}int query(int l, int r) {    int len = r-l+1;    int k = 0;    while((1<<(k+1)) <= len) k++;    return lca(rmq[l][k], rmq[r-(1<<k)+1][k]);}    int main() {    while(scanf("%d", &n) != EOF) {        init();        int u, v;        for(int i = 0;i < n-1; i++) {            scanf("%d%d", &u, &v);            addEdge(u, v);            addEdge(v, u);        }        dfs(1, -1, 0);        for(int i = 0;i+1<20; i++) {            for(int u = 1;u <= n; u++) {                if(par[u][i] < 0) par[u][i+1] = -1;                else par[u][i+1] = par[par[u][i]][i];            }        }        for(int i = 1;i <= n; i++) rmq[i][0] = i;        for(int k = 0;k < 20; k++)             for(int i = 1;i <= n; i++) {                if(i+(1<<(k+1))-1 > n) break;                rmq[i][k+1] = lca(rmq[i][k], rmq[i+(1<<k)][k]);                //cout<<rmq[i][k+1]<<endl;            }        //cout<<rmq[2][2]<<endl;        scanf("%d", &q);        int l, r;        while(q--) {            scanf("%d%d", &l, &r);            printf("%d\n", query(l, r));        }    }    return 0;}

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