Hihocoder 最近公用祖先三 在线LCA

在线的LCA算法，dfs遍历整棵树，对于每个点出现的时候都插入到数组中，然后查询两个点的lca就是两个点在数组中最后出现位置间的dep值最小的点，就转化为链上的RMQ问题了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <cmath>
#include <climits>

using namespace std;
const int maxn = 1e5 + 10;
int head[maxn], nxt[maxn << 1], v[maxn << 1];
int rpos[maxn], n, Q, cnt, ecnt;
map<string, int> mp;
map<int, string> smp;
char name1[1024], name2[1024];

struct Node {
	int dep, id;
	bool operator < (const Node &x) const {
		return dep < x.dep;
	}
};

Node val[maxn << 1], minv[maxn << 1][30];

void adde(int uu, int vv) {
	v[ecnt] = vv; nxt[ecnt] = head[uu]; head[uu] = ecnt++;
}

int getid(char *str) {
	if(mp.count(str) == 0) {
		int mpz = mp.size();
		mp[str] = mpz + 1;
		smp[mpz + 1] = str;
		return mpz + 1;
	}
	return mp[str];
}

void dfs(int now, int fa, int dep) {
	val[++cnt].dep = dep; val[cnt].id = now;
	for(int i = head[now]; ~i; i = nxt[i]) if(v[i] != fa) {
		dfs(v[i], now, dep + 1);
		val[++cnt].dep = dep; val[cnt].id = now;
	}
	rpos[now] = cnt;
}

void initRMQ() {
	for(int i = 1; i <= cnt; i++) {
		minv[i][0] = val[i];
	}
	for(int j = 1; (1 << j) <= cnt; j++) {
		for(int i = 1; i + (1 << j) - 1 <= cnt; i++) {
			minv[i][j] = min(minv[i][j - 1], minv[i + (1 << (j - 1))][j - 1]);
		}
	}
}

Node query(int l, int r) {
	int k = 0;
	while((1 << (k + 1)) <= r - l + 1) k++;
	return min(minv[l][k], minv[r - (1 << k) + 1][k]);
}

int main() {
	memset(head, -1, sizeof(head));
	scanf("%d", &n);
	for(int i = 1; i <= n; i++) {
		scanf("%s%s", name1, name2);
		int a = getid(name1), b = getid(name2);
		adde(a, b); adde(b, a);
	}
	dfs(1, -1, 0);
	initRMQ();
	scanf("%d", &Q);
	while(Q--) {
		scanf("%s%s", name1, name2);
		int a = getid(name1), b = getid(name2);
		a = rpos[a]; b = rpos[b];
		if(a > b) swap(a, b);
		Node ret = query(a, b);
		puts(smp[ret.id].c_str());
	}
	return 0;
}