HDU 5266 pog loves szh III（区间LCA）

题目链接 pog loves szh III

题意就是  求一个区间所有点的$LCA$。

我们把$1$到$n$的$DFS$序全部求出来……然后设$i$的$DFS$序为$c[i]$，$pc[i]$为$c[i]$的反函数。

区间的$LCA$其实就是，$DFS$序最大和最小的两个点的$LCA$。

（话说$2017$女生赛里面有一题要用的结论和这题的差不多）

然后求出区间的$DFS$序最大值$x$和最小值$y$。

然后求一下$LCA(pc[x],pc[y])$即可。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)

typedef long long LL;

const int N = 300010;
const int A = 21;

int c[N];
int deep[N];
int f[N][A], g[N][A], st[N][A];
vector <int> v[N];
int ti;
int n, q;
int pc[N];

void ST(){
	rep(i, 1, n) f[i][0] = c[i];
	rep(j, 1, 20) rep(i, 1, n)
		if ((i + (1 << j) - 1) <= n) f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);

	rep(i, 1, n) g[i][0] = c[i];
	rep(j, 1, 20) rep(i, 1, n)
		if ((i + (1 << j) - 1) <= n) g[i][j] = max(g[i][j - 1], g[i + (1 << (j - 1))][j - 1]);
}

inline int solvemin(int l, int r){
	int k = (int)log2((double)(r - l + 1));
	return min(f[l][k], f[r - (1 << k) + 1][k]);

}

inline int solvemax(int l, int r){
	int k = (int)log2((double)(r - l + 1));
	return max(g[l][k], g[r - (1 << k) + 1][k]);
}


void dfs(int x, int fa, int dep){
	c[x] = ++ti; pc[ti] = x;
	deep[x] = dep;
	if (fa){
		st[x][0] = fa;
		for (int i = 0; st[st[x][i]][i]; ++i) st[x][i + 1] = st[st[x][i]][i];
	}

	for (auto u : v[x]){
		if (u == fa) continue;
		dfs(u, x, dep + 1);
	}
}

int LCA(int a, int b){
	if (deep[a] < deep[b]) swap(a, b);
	for (int i = 0,  delta = deep[a] - deep[b]; delta; delta >>= 1, ++i) if (delta & 1) a = st[a][i];
	if (a == b) return a;
	dec(i, 19, 0) if (st[a][i] != st[b][i]) a = st[a][i], b = st[b][i];
	return st[a][0];
}

int main(){

	while (~scanf("%d", &n)){
		rep(i, 0, n + 1) v[i].clear();
		memset(c, 0, sizeof c);
		ti = 0;
		rep(i, 2, n){
			int x, y;
			scanf("%d%d", &x, &y);
			v[x].push_back(y);
			v[y].push_back(x);
		}

		memset(st, 0, sizeof st);
		memset(f, 0, sizeof f);
		memset(g, 0, sizeof g);
		dfs(1, 0, 0);
		ST();
		for (scanf("%d", &q); q--; ){
			int x, y;
			scanf("%d%d", &x, &y);
			printf("%d\n", LCA(pc[solvemax(x, y)], pc[solvemin(x, y)]));

		}
	}


	return 0;
}