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  • 洛谷P2590 [ZJOI2008]树的统计 题解 树链剖分+线段树

    题目链接:https://www.luogu.org/problem/P2590
    树链剖分模板题。
    剖分过程要用到如下7个值:

    • fa[u]:u的父节点编号;
    • dep[u]:u的深度;
    • size[u]:u为根的子树中节点总数;
    • son[u]:u的重儿子;
    • top[u]:u所在的重链的顶部节点;
    • seg[u]:u在线段树中的位置;
    • rev[u]:seg的倒置,即rev[seg[u]] == u

    然后套线段树模板实现区间最值、区间和,及单点更新操作。

    实现代码如下:

    #include <bits/stdc++.h>
    using namespace std;
    #define INF (1<<29)
    const int maxn = 30030;
    int fa[maxn],
    	dep[maxn],
    	size[maxn],
    	son[maxn],
    	top[maxn],
    	seg[maxn], seg_cnt,
    	rev[maxn],
    	n, w[maxn], maxv[maxn<<2], sumv[maxn<<2];
    vector<int> g[maxn];
    void dfs1(int u, int p) {
    	size[u] = 1;
    	for (vector<int>::iterator it = g[u].begin(); it != g[u].end(); it ++) {
    		int v = (*it);
    		if (v == p) continue;
    		fa[v] = u;
    		dep[v] = dep[u] + 1;
    		dfs1(v, u);
    		size[u] += size[v];
    		if (size[v] >size[son[u]]) son[u] = v;
    	}
    }
    void dfs2(int u, int tp) {
    	seg[u] = ++seg_cnt;
    	rev[seg_cnt] = u;
    	top[u] = tp;
    	if (son[u]) dfs2(son[u], tp);
    	for (vector<int>::iterator it = g[u].begin(); it != g[u].end(); it ++) {
    		int v = (*it);
    		if (v == fa[u] || v == son[u]) continue;
    		dfs2(v, v);
    	}
    }
    #define lson l, mid, rt<<1
    #define rson mid+1, r, rt<<1|1
    void push_up(int rt) {
    	sumv[rt] = sumv[rt<<1] +sumv[rt<<1|1];
    	maxv[rt] = max(maxv[rt<<1], maxv[rt<<1|1]);
    }
    void build(int l, int r, int rt) {
    	int mid = (l + r) / 2;
    	if (l == r) {
    		sumv[rt] = maxv[rt] = w[rev[l]];
    		return;
    	}
    	build(lson); build(rson);
    	push_up(rt);
    }
    void update(int p, int v, int l, int r, int rt) {
    	if (l == r) {
    		sumv[rt] = maxv[rt] = v;
    		return;
    	}
    	int mid = (l + r) / 2;
    	if (p <= mid) update(p, v, lson);
    	else update(p, v, rson);
    	push_up(rt);
    }
    int query_max(int L, int R, int l, int r, int rt) {
    	if (L <= l && r <= R) return maxv[rt];
    	int mid = (l + r) / 2, tmp = -INF;
    	if (L <= mid) tmp = max(tmp, query_max(L, R, lson));
    	if (R > mid) tmp = max(tmp, query_max(L, R, rson));
    	return tmp;
    }
    int query_sum(int L, int R, int l, int r, int rt) {
    	if (L <= l && r <= R) return sumv[rt];
    	int mid = (l + r) / 2, tmp = 0;
    	if (L <= mid) tmp += query_sum(L, R, lson);
    	if (R > mid) tmp += query_sum(L, R, rson);
    	return tmp;
    }
    int ask_max(int u, int v) {
    	int res = -INF;
    	while (top[u] != top[v]) {
    		if (dep[top[u]] < dep[top[v]]) swap(u, v);
    		res = max(res, query_max(seg[top[u]], seg[u], 1, n, 1));
    		u = fa[top[u]];
    	}
    	if (dep[u] < dep[v]) swap(u, v);
    	res = max(res, query_max(seg[v], seg[u], 1, n, 1));
    	return res;
    }
    int ask_sum(int u, int v) {
    	int res = 0;
    	while (top[u] != top[v]) {
    		if (dep[top[u]] < dep[top[v]]) swap(u, v);
    		res += query_sum(seg[top[u]], seg[u], 1, n, 1);
    		u = fa[top[u]];
    	}
    	if (dep[u] < dep[v]) swap(u, v);
    	res += query_sum(seg[v], seg[u], 1, n, 1);
    	return res;
    }
    int m;
    string s;
    int main() {
    	cin >> n;
    	for (int i = 1; i < n; i ++) {
    		int u, v;
    		cin >> u >> v;
    		g[u].push_back(v);
    		g[v].push_back(u);
    	}
    	for (int i = 1; i <= n; i ++) cin >> w[i];
    	dep[1] = fa[1] = 1;
    	dfs1(1, -1);
    	dfs2(1, 1);
    	build(1, n, 1);
    	cin >> m;
    	while (m --) {
    		int u, v;
    		cin >> s >> u >> v;
    		if (s == "CHANGE") update(seg[u], v, 1, n, 1);
    		else if (s == "QMAX") cout << ask_max(u, v) << endl;
    		else cout << ask_sum(u, v) << endl;
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/codedecision/p/11769437.html
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