树的统计

金牌导航 树链剖分-1

题目大意

给出一棵树，让你做若干操作，操作如下：
1.修改一个节点的值
2.查询两个节点之间路径的最大值
3.查询两个节点之间路径的和

输入样例

4
1 2
2 3
4 1
4 2 1 3
12
QMAX 3 4
QMAX 3 3
QMAX 3 2
QMAX 2 3
QSUM 3 4
QSUM 2 1
CHANGE 1 5
QMAX 3 4
CHANGE 3 6
QMAX 3 4
QMAX 2 4
QSUM 3 4

输出样例

4
1
2
2
10
6
5
6
5
16

数据范围

$1⩽N⩽3\times 10^4,0⩽q⩽2\times 10^5,-3\times 10^4⩽s_i⩽3\times 10^4$

解题思路

树链剖分，然后用线段树维护重链，每个节点维护最大值和权值和

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 30030
using namespace std;
int n, m, x, y, w, tot, ansmax, anssum;
int a[N], s[N<<4], v[N], fa[N], hs[N], han[N], dfn[N], dep[N], size[N], head[N], maxx[N<<4];
string str;
struct rec
{int to, next;
}e[N<<1];
void add(int x, int y)
{e[++tot].to = y;e[tot].next = head[x];head[x] = tot;
}
void dfs1(int x)//找重儿子
{size[x] = 1;for (int i = head[x]; i; i = e[i].next)if (e[i].to != fa[x]){fa[e[i].to] = x;dep[e[i].to] = dep[x] + 1;dfs1(e[i].to);size[x] += size[e[i].to];if (size[e[i].to] > size[hs[x]]) hs[x] = e[i].to;}return;
}
void dfs2(int x)
{dfn[x] = ++w;v[w] = x;if (hs[x]){han[hs[x]] = han[x];//重祖先dfs2(hs[x]);}for (int i = head[x]; i; i = e[i].next)if (e[i].to != fa[x] && e[i].to != hs[x]){han[e[i].to] = e[i].to;dfs2(e[i].to);}
}
void up(int x)
{s[x] = s[x * 2] + s[x * 2 + 1];maxx[x] = max(maxx[x * 2], maxx[x * 2 + 1]);return;
}
void build(int now, int l, int r)//线段树维护
{if (l == r){maxx[now] = s[now] = a[v[l]];return;}int mid = (l + r) >> 1;build(now * 2, l, mid);build(now * 2 + 1, mid + 1, r);up(now);return;
}
void change(int x, int y, int now, int l, int r)
{if (l == r){maxx[now] = s[now] = y;return;}int mid = (l + r) >> 1;if (x <= mid) change(x, y, now * 2, l, mid);else change(x, y, now * 2 + 1, mid + 1, r);up(now);return;
}
void ask(int now, int ql, int qr, int l, int r)
{if (l == ql && r == qr){ansmax = max(ansmax, maxx[now]);anssum += s[now];return;}int mid = (l + r) >> 1;if (qr <= mid) {ask(now * 2, ql, qr, l, mid); return;}if (ql > mid) {ask(now * 2 + 1, ql, qr, mid + 1, r); return;}ask(now * 2, ql, mid, l, mid);ask(now * 2 + 1, mid + 1, qr, mid + 1, r);return;
}
void askk(int x, int y)//计算路径长度
{anssum = 0;ansmax = -N;while(han[x] != han[y]){if (dep[han[x]] < dep[han[y]]) swap(x, y);ask(1, dfn[han[x]], dfn[x], 1, n);x = fa[han[x]];}if (dep[x] < dep[y]) swap(x, y);ask(1, dfn[y], dfn[x], 1, n);return;
}
int main()
{scanf("%d", &n);for (int i = 1; i < n; ++i){scanf("%d%d", &x, &y);add(x, y);add(y, x);}for (int i = 1; i <= n; ++i)scanf("%d", &a[i]);fa[1] = 1;han[1] = 1;dfs1(1);dfs2(1);build(1, 1, n);scanf("%d", &m);while(m--){cin>>str;scanf("%d%d", &x, &y);if (str == "CHANGE"){change(dfn[x], y, 1, 1, n);}else{askk(x, y);if (str == "QSUM") printf("%d\n", anssum);else if (str == "QMAX") printf("%d\n", ansmax);}}return 0;
}