【ZJOI 2008】 树的统计 - 树链剖分

题目描述

一棵树上有n个节点，编号分别为1到n，每个节点都有一个权值w。

我们将以下面的形式来要求你对这棵树完成一些操作：

I. CHANGE u t : 把结点u的权值改为t

II. QMAX u v: 询问从点u到点v的路径上的节点的最大权值

III. QSUM u v: 询问从点u到点v的路径上的节点的权值和

注意：从点u到点v的路径上的节点包括u和v本身

思路

树剖板子　　

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <vector>
#include <string>
using namespace std;
const int maxn = 100000 + 10;
int cnt = 0,map[maxn],val[maxn],num[maxn],size[maxn],father[maxn],son[maxn],top[maxn],dep[maxn],w[maxn];
int n,m,r,p;
vector<int> edges[maxn];
inline void dfs1(int now,int f) {
    size[now] = 1;
    father[now] = f;
    dep[now] = dep[father[now]]+1;
    for (size_t i = 0;i < edges[now].size();i++)
        if (edges[now][i] != f) {
            dfs1(edges[now][i],now);
            size[now] += size[edges[now][i]];
            if (size[son[now]] < size[edges[now][i]] || !son[now]) son[now] = edges[now][i];
        }
}
inline void dfs2(int now,int ntop) {
    top[now] = ntop;
    num[now] = ++cnt;
    map[num[now]] = now;
    if (son[now]) dfs2(son[now],ntop);
    for (size_t i = 0;i < edges[now].size();i++)
        if (edges[now][i] != father[now] && edges[now][i] != son[now]) dfs2(edges[now][i],edges[now][i]);
}
struct seg { int maxv,sum,mark,l,r; } tree[maxn*4];
inline void pushup(int root) {
    tree[root].sum = tree[root<<1].sum+tree[root<<1|1].sum;
    tree[root].maxv = max(tree[root<<1].maxv,tree[root<<1|1].maxv);
}
inline void BuildTree(int l,int r,int root) {
    tree[root].l = l;
    tree[root].r = r;
    tree[root].mark = 0;
    if (l == r) {
        tree[root].sum = val[map[l]];
        tree[root].maxv = val[map[l]];
        return;
    }
    int mid = l+r>>1;
    BuildTree(l,mid,root<<1);
    BuildTree(mid+1,r,root<<1|1);
    pushup(root);
}
inline void Update(int l,int r,int num,int root,int x) {
    if (l == r) {
        tree[root].sum = x;
        tree[root].maxv = x;
        return;
    }
    int mid = l+r>>1;
    if (num <= mid) Update(l,mid,num,root<<1,x);
    else Update(mid+1,r,num,root<<1|1,x);
    pushup(root);
}
inline int QuerySum(int l,int r,int ql,int qr,int root) {
    if (ql > r || qr < l) return 0;
    if (ql <= l && qr >= r) return tree[root].sum;
    int mid = l+r>>1;
    return QuerySum(l,mid,ql,qr,root<<1)+QuerySum(mid+1,r,ql,qr,root<<1|1);
}
inline int QueryMax(int l,int r,int ql,int qr,int root) {
    if (ql > r || qr < l) return -999999999;
    if (ql <= l && qr >= r) return tree[root].maxv;
    int mid = l+r>>1;
    return max(QueryMax(l,mid,ql,qr,root<<1),QueryMax(mid+1,r,ql,qr,root<<1|1));
}
inline int QuerySumEdges(int u,int v) {
    int topu = top[u];
    int topv = top[v];
    int sum = 0;
    while (topu != topv) {
        if (dep[topu] < dep[topv]) {
            swap(topu,topv);
            swap(u,v);
        }
        sum += QuerySum(1,cnt,num[topu],num[u],1);
        u = father[topu];
        topu = top[u];
    }
    if (dep[u] > dep[v]) swap(u,v);
    return sum+QuerySum(1,cnt,num[u],num[v],1);
}
inline int QueryMaxEdges(int u,int v) {
    int topu = top[u];
    int topv = top[v];
    int Max = -999999999;
    while (topu != topv) {
        if (dep[topu] < dep[topv]) {
            swap(topu,topv);
            swap(u,v);
        }
        Max = max(Max,QueryMax(1,cnt,num[topu],num[u],1));
        u = father[topu];
        topu = top[u];
    }
    if (dep[u] > dep[v]) swap(u,v);
    return max(Max,QueryMax(1,cnt,num[u],num[v],1));
}
int main() {
    ios :: sync_with_stdio(false);
    cin >> n;
    for (int i = 1,u,v;i < n;i++) {
        cin >> u >> v;
        edges[u].push_back(v);
        edges[v].push_back(u);
    }
    for (int i = 1;i <= n;i++) cin >> val[i];
    dfs1(1,0);
    dfs2(1,1);
    BuildTree(1,cnt,1);
    cin >> m;
    while (m--) {
        string dispose;
        int u,v;
        cin >> dispose >> u >> v;
        if (dispose == "CHANGE") Update(1,cnt,num[u],1,v);
        else if (dispose == "QSUM") cout << QuerySumEdges(u,v) << endl;
        else cout << QueryMaxEdges(u,v) << endl;
    }
    return 0;
}