洛谷P3178[HAOI]2015 树上操作

题目

树剖裸题，这个题更可以深刻的理解树剖中把树上的节点转换为区间的思想。

要注意在区间上连续的节点，一定是在一棵子树中。

#include <bits/stdc++.h>
#define int long long
#define ls left, mid, root << 1
#define rs mid + 1, right, root << 1 | 1
#define N 600100
using namespace std;
int n, m, rot, mod, tot, cnt;
int data[N], id[N], dep[N], size[N], lin[N], ans[N * 8], lazy[N * 8], dp[N], fa[N], top[N], son[N];
struct edg {
	int to, nex;
} e[N];
inline void add(int f, int t)
{
	e[++cnt].to = t;
	e[cnt].nex = lin[f];
	lin[f] = cnt;
}
inline void pushup(int root)
{
 	ans[root] = (ans[root << 1] + ans[root << 1 | 1]);
}
inline void pushdown(int root, int left, int right)
{
  	int mid = (left + right) >> 1;
   	if (lazy[root])
   	{
		ans[root << 1] += (mid - left + 1) * lazy[root];
		ans[root << 1];
		ans[root << 1 | 1] += (right - mid) * lazy[root];
		ans[root << 1 | 1];
		lazy[root << 1] += lazy[root];
		lazy[root << 1 | 1] += lazy[root];
		lazy[root] = 0;
	}
}
void build(int left, int right, int root)
{
	if (left == right)
	{
		ans[root] = dp[left], ans[root];
		return;
	}
	int mid = (left + right) >> 1;
	build(ls), build(rs);
	pushup(root);
}
inline void update(int left, int right, int root, int add, int ql, int qr)
{
	if (left >= ql && right <= qr)
	{
		ans[root] += (right - left + 1) * add;
		lazy[root] += add;
		ans[root];
		return;
	}
	int mid = (left + right) >> 1;
	pushdown(root, left, right);//线段树的pushdown操作是为了弥补之前没向下传递标记的坑。
	if (ql <= mid)
		update(ls, add, ql, qr);
	if (qr > mid)
		update(rs, add, ql, qr);
	pushup(root);
}
inline int query(int left, int right, int root, int ql, int qr)
{
	int res = 0;
	if (left >= ql && right <= qr)
		return ans[root];
	int mid = (left + right) >> 1;
	pushdown(root, left, right);
	if (ql <= mid)
		res = ( res + query(ls, ql, qr) );
	if (qr > mid)
		res = res + query(rs, ql, qr) ;
	return res;
}
void dfs1(int now, int f, int de)
{
	fa[now] = f, dep[now] = de, size[now] = 1;
	int maxsize = -1;
	for (int i = lin[now]; i; i = e[i].nex)
	{
		if (e[i].to == f) continue;
		dfs1(e[i].to, now, de + 1);
		size[now] += size[e[i].to];
		if (size[e[i].to] > maxsize)
		{
			maxsize = size[e[i].to];
			son[now] = e[i].to;
		}
	}
}
void ulca(int x, int y, int z)
{
	while (top[x] != top[y])
	{
		if (dep[top[x]] < dep[top[y]])
			swap(x, y);
		update(1, n, 1, z, id[top[x]], id[x]);
		x = fa[top[x]];
	}
	if (dep[x] > dep[y])
		swap(x, y);
	update(1, n, 1, z, id[x], id[y]);
}
int qlca(int x, int y)
{
	int res = 0;
	while (top[x] != top[y])
	{
		if (dep[top[x]] < dep[top[y]])
			swap(x, y);
		res = ( res + query(1, n, 1, id[top[x]], id[x]) );
		x = fa[top[x]];
	}
	if (dep[x] > dep[y])
		swap(x, y);
	res = ( res + query(1, n, 1, id[x], id[y]) );
	return res;
}
void upd(int x, int y)
{
	update(1, n, 1, y, id[x], id[x] + size[x] - 1);
}
int que(int x)
{
	return query(1, n, 1, id[x], id[x] + size[x] - 1);
}
void dfs2(int now, int t)
{
	top[now] = t;
	dp[++tot] = data[now];
	id[now] = tot;
	if (!son[now])
		return;
	dfs2(son[now], t);
	for (int i = lin[now]; i; i = e[i].nex)
	{
		int to = e[i].to;
		if (to != fa[now] && to != son[now])
			dfs2(to, to);
	}
}
signed main()
{
	scanf("%lld%lld", &n, &m);
	rot = 1;
	for (int i = 1; i <= n; i++)
			scanf("%lld", &data[i]);
	for (int i = 1; i < n; i++)
	{
		int a, b;
		scanf("%lld%lld", &a, &b);
		add(a, b);
		add(b, a);
	}
	dfs1(rot, 0, 1);
	dfs2(rot, rot);
	build(1, n, 1);
	for (int i = 1; i <= m; i++)
	{
		int flag;
		scanf("%lld", &flag);
		if (flag == 1)
		{
			int x, a;
			scanf("%lld%lld", &x, &a);
			update(1, n, 1, a, id[x], id[x]);
		}
		if (flag == 2)
		{
			int x, a;
			scanf("%lld%lld", &x, &a);
			update(1, n, 1, a, id[x],  id[x] + size[x] - 1);
		}
		if (flag == 3)
		{
			int a;
			scanf("%lld", &a);
			printf("%lld\n", qlca(1, a));//不能写成id[1],id[a]因为id子树之间是连续的，所以a到1之间并不是连续边号。
		}
	}
	return 0;
}
/*
5 5 2 24
7 3 7 8 0
1 2
1 5
3 1
4 1
3 4 2
3 2 2
4 5
1 5 1 3
2 1 3
*/