2019 Nanchang Network Game J. Distance on the tree

Copyright: https: //blog.csdn.net/qq_40831340 welcome treatise https://blog.csdn.net/qq_40831340/article/details/90739880

This problem wa rewrite it again after what I first wrote of dog feces ah

Topic: https://nanti.jisuanke.com/t/38229

As shown below

for each tree node to him to the back of the President of each chain to build the tree
so long as lca can run fast check interval k big

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 100000 + 10;

int head[maxn], depth[maxn], cnt;
int shu;
int dis[maxn << 1];
int nxt[maxn << 1], to[maxn << 1];

int fa[maxn][30], lg[maxn];
int n, m, k, q;

void ade(int a, int b, int c) {
	to[++cnt] = b;
	nxt[cnt] = head[a];
	dis[cnt] = c;
	head[a] = cnt;
}

struct node {
	int lc, rc;
	int val;
} tree[maxn * 20];

int tot, root[maxn];
int b[maxn];

int build(int l, int r) {
	int p = ++ tot;
	if(l == r) {
		tree[p].val = 0;
		return p;
	}
	int mid = l + r >> 1;
	tree[p].lc = build(l, mid);
	tree[p].rc = build(mid + 1, r);
	tree[p].val = tree[tree[p].lc].val + tree[tree[p].rc].val;
	return p;
}

int ins(int now, int l, int r, int pos, int val) {
	int p = ++ tot;
	tree[p] = tree[now];
	if(l == r) {
		tree[p].val += val ;
		return p;
	}
	int mid = l + r >> 1;
	if(pos <= mid)
		tree[p].lc = ins(tree[now].lc, l, mid, pos, val);
	else
		tree[p].rc = ins(tree[now].rc, mid + 1, r, pos, val);
	tree[p].val = tree[tree[p].lc].val + tree[tree[p].rc].val;
	return p;
}

int ask(int p, int q, int l, int r, int L, int R) {
	if(R < L) return 0;
	if(L <= l && r <= R)
		return tree[p].val - tree[q].val;

	int mid = l + r >> 1;
	int res = 0;
	if(L <= mid)
		res += ask(tree[p].lc, tree[q].lc, l, mid, L, R);
	if(R > mid)
		res += ask(tree[p].rc, tree[q].rc, mid + 1, r, L, R);
	return res;
}

int getid(int x) {
	int res = upper_bound(b + 1, b + shu, x) - b - 1;
	return res;
}

void dfs(int x, int pre) {
	depth[x] = depth[pre] + 1;
	fa[x][0] = pre;
	for(int i = 1; (1 << i) <= depth[x]; i++)
		fa[x][i] = fa[fa[x][i - 1]][i - 1];
	for(int i = head[x]; i; i = nxt[i])
		if(to[i] != pre) {
			root[to[i]] = ins(root[x], 1, shu, getid(dis[i]), 1);
			dfs(to[i], x);
		}
}

int lca(int x, int y) {
	if(depth[x] < depth[y])
		swap(x, y);
	while(depth[x] > depth[y])
		x = fa[x][lg[depth[x] - depth[y]] - 1];
	if(x == y)
		return x;
	for(int k = lg[depth[x]] - 1; k >= 0; k--) {
		if(fa[x][k] != fa[y][k]) {
			x = fa[x][k], y = fa[y][k];
		}
	}
	return fa[x][0];
}

void lsh() {
	sort(b + 1, b + n);
	shu = unique(b + 1, b + n) - b;
}

int main() {
	for(int i = 1; i < maxn; i++)
		lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);

	scanf("%d %d", &n, &m);
	for(int i = 1; i < n; i ++ ) {
		int st, ed, dis;
		cin >> st >> ed >> dis;
		ade(st, ed, dis),ade(ed, st, dis);
		b[i] = dis;
	}

	lsh();

	root[1] = build(1, shu);

	dfs(1,0);

	for(int i = 1; i <= m; i ++) {
		int x, y, k;
		scanf( "%d %d %d" , &x, &y, &k);
		k = getid(k);
		int LCA = lca(x, y);
		printf( "%d\n",\
		        ask(root[y], root[LCA], 1, shu, 1, k)\
		        + ask(root[x], root[LCA], 1, shu, 1, k)\
		      );
	}
	return 0;
}