SPOJ - QTREE Query on a tree 树链剖分+线段树形式

You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.

We will ask you to perfrom some instructions of the following form:

• CHANGE i ti : change the cost of the i-th edge to ti
  or
• QUERY a b : ask for the maximum edge cost on the path from node a to node b

Input

The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.

For each test case:

• In the first line there is an integer N (N <= 10000),
• In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 1000000),
• The next lines contain instructions "CHANGE i ti" or "QUERY a b",
• The end of each test case is signified by the string "DONE".

There is one blank line between successive tests.

Output

For each "QUERY" operation, write one integer representing its result.

Example

Input:
1

3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE

Output:
1
3

#include<bits/stdc++.h>
using namespace std;
const int maxn = 50010;
struct edge {
	int v, ne;
}ed[maxn * 2];
int head[maxn], cnt, n;
int fa[maxn], son[maxn], num[maxn], deep[maxn];
int p[maxn], fp[maxn], top[maxn];
int pos;
void init() {
	cnt = 0;
	memset(head, -1, sizeof head);
	pos = 1;
	memset(son, -1, sizeof son);
}
void addedge(int u, int v) {
	ed[cnt].v = v;
	ed[cnt].ne = head[u];
	head[u] = cnt++;
}
void dfs1(int u, int pre, int d) {
	deep[u] = d;
	fa[u] = pre;
	num[u] = 1;
	for (int i = head[u]; ~i; i = ed[i].ne) {
		int v = ed[i].v;
		if (v == pre)continue;
		dfs1(v, u, d + 1);
		num[u] += num[v];
		if (son[u] == -1 || num[v] > num[son[u]])
			son[u] = v;
	}
}
void getpos(int u, int sp) {
	top[u] = sp;
	p[u] = pos++;
	fp[p[u]] = u;
	if (son[u] == -1)return;
	getpos(son[u], sp);
	for (int i = head[u]; ~i; i = ed[i].ne) {
		int v = ed[i].v;
		if (v != son[u] && v != fa[u]) {
			getpos(v, v);
		}
	}
}
struct node {
	int l, r;
	int Max;
}c[maxn * 4];
void build(int rt, int l, int r) {
	c[rt].l = l; c[rt].r = r;
	c[rt].Max = 0;
	if (l == r)return;
	int mid = (l + r) >> 1;
	build(rt << 1, l, mid);
	build((rt << 1) | 1, mid + 1, r);
}
void pushup(int rt) {
	c[rt].Max = max(c[rt << 1].Max, c[(rt << 1) | 1].Max);
}
void update(int rt, int k, int val) {
	if (c[rt].l == k&&c[rt].r == k) {
		c[rt].Max = val;
		return;
	}
	int mid = (c[rt].l + c[rt].r) >> 1;
	if (k <= mid)update(rt << 1, k, val);
	else update(((rt << 1) | 1), k, val);
	pushup(rt);
}
int query(int rt, int l, int r) {
	if (c[rt].l == l&&c[rt].r == r)
		return c[rt].Max;
	int mid = (c[rt].l + c[rt].r) >> 1;
	if (r <= mid)return query(rt << 1, l, r);
	else if (l > mid)return query((rt << 1) | 1, l, r);
	else return max(query(rt << 1, l, mid), query((rt << 1) | 1, mid + 1, r));
}
//查询u->v的最大值
int find(int u, int v) {
	int f1 = top[u], f2 = top[v];
	int tmp = 0;
	while (f1 != f2) {
		if (deep[f1] < deep[f2]) {
			swap(f1, f2);
			swap(u, v);
		}
		tmp = max(tmp, query(1, p[f1], p[u]));
		u = fa[f1]; f1 = top[u];
	}
	if (u == v)return tmp;
	if (deep[u] > deep[v])swap(u, v);
	return max(tmp, query(1, p[son[u]], p[v]));
}
int e[maxn][3];
int main() {
	int te, n;
	ios::sync_with_stdio(0);
	cin >> te;
	while (te--) {
		init();
		cin >> n;
		for (int i = 1; i <= n - 1; i++) {
			cin >> e[i][0] >> e[i][1] >> e[i][2];
			addedge(e[i][0], e[i][1]);
			addedge(e[i][1], e[i][0]);
		}
		dfs1(1, 0, 0);
		getpos(1, 1);
		build(1, 1, pos);
		for (int i = 1; i <= n - 1; i++) {
			if (deep[e[i][0]] < deep[e[i][1]]) {
				swap(e[i][0], e[i][1]);
			}
			update(1, p[e[i][0]], e[i][2]);
		}
		string que;
		while (cin >> que) {
			int a, b;
			if (que[0] == 'D')break;
			else if (que[0] == 'C') {
				cin >> a >> b;
				update(1, p[e[a][0]], b);
			}
			else {
				cin >> a >> b;
				cout << find(a, b) << endl;
			}
		}
	}
	return 0;
}