Query on a tree 树链剖分 [模板]

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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 100005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == ‘-‘) f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/
struct node {
	int to, nxt;
}e[maxn];
int head[maxn], tot;
int top[maxn];// top[v]表示v所在的重链的顶点
int fa[maxn];
int dep[maxn];
int num[maxn];// num[v]表示以v为根的子树大小
int p[maxn];// p[v]表示v与其父亲节点在线段树的位置
int fp[maxn];
int son[maxn];// 重儿子
int pos;
int n;

void init() {
	tot = 0; memset(head, -1, sizeof(head));
	pos = 1; memset(son, -1, sizeof(son));
}
void addedge(int u, int v) {
	e[tot].to = v; e[tot].nxt = head[u]; head[u] = tot++;
}

void dfs1(int u, int pre, int d) {
	dep[u] = d; fa[u] = pre; num[u] = 1;
	for (int i = head[u]; i != -1; i = e[i].nxt) {
		int v = e[i].to;
		if (v != pre) {
			dfs1(v, u, d + 1);
			num[u] += num[v];
			if (son[u] == -1 || num[v] > num[son[u]])son[u] = v;
		}
	}
}

void dfs2(int u, int sp) {
	top[u] = sp;
	if (son[u] != -1) {
		p[u] = pos++;
		fp[p[u]] = u;
		dfs2(son[u], sp);
	}
	else {
		p[u] = pos++; fp[p[u]] = u; return;
	}
	for (int i = head[u]; i != -1; i = e[i].nxt) {
		int v = e[i].to;
		if (v != son[u] && v != fa[u])dfs2(v, v);
	}
}

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int maxx[maxn];
int val[maxn];
void pushup(int rt) {
	maxx[rt] = max(maxx[rt << 1], maxx[rt << 1 | 1]);
}
void build(int l, int r, int rt) {
	if (l == r) {
		maxx[rt] = val[l]; return;
	}
	int m = (l + r) >> 1;
	build(lson); build(rson); pushup(rt);
}
void upd(int p, int x, int l, int r, int rt) {
	if (l == r) {
		maxx[rt] = x; return;
	}
	int m = (l + r) >> 1;
	if (p <= m)upd(p, x, lson);
	else upd(p, x, rson);
	pushup(rt);
}
int query(int L, int R, int l, int r, int rt) {
	if (L <= l && r <= R) {
		return maxx[rt];
	}
	int m = (l + r) >> 1;
	int ans = 0;
	if (L <= m)ans = max(ans, query(L, R, lson));
	if (m < R)ans = max(ans, query(L, R, rson));
	return ans;
}

int Find(int u, int v) {
	int f1 = top[u], f2 = top[v];
	int tmp = 0;
	while (f1 != f2) {
		if (dep[f1] < dep[f2]) {
			swap(f1, f2); swap(u, v);
		}
		tmp = max(tmp, query(p[f1], p[u], 1, n, 1));
		u = fa[f1]; f1 = top[u];
	}
	if (u == v)return tmp;
	if (dep[u] > dep[v])swap(u, v);
	return max(tmp, query(p[son[u]], p[v], 1, n, 1));
}
int ed[maxn][3];

int main()
{
	//	ios::sync_with_stdio(0);
	int t; t = rd();
	while (t--) {
		init(); n = rd();
	//	getchar();
		for (int i = 0; i < n - 1; i++) {
			ed[i][0] = rd(); ed[i][1] = rd(); ed[i][2] = rd();
			addedge(ed[i][0], ed[i][1]);
			addedge(ed[i][1], ed[i][0]);
		}
		dfs1(1, 0, 0); dfs2(1, 1);
		for (int i = 0; i < n - 1; i++) {
			if (dep[ed[i][0]] < dep[ed[i][1]]) {
				swap(ed[i][0], ed[i][1]);
			}
			val[p[ed[i][0]]] = ed[i][2];
		}
		build(1, n, 1);
		char opt[10];
		while (scanf("%s",opt)) {
			if (opt[0] == ‘D‘)break;
			int u, v; u = rd(); v = rd();
			if (opt[0] == ‘Q‘) {
				printf("%d\n", Find(u, v));
			}
			else upd(p[ed[u - 1][0]], v, 1, n, 1);
		}
	}
	return 0;
}

原文地址：https://www.cnblogs.com/zxyqzy/p/10354449.html