[洛谷P3178][HAOI2015]树上操作

题目大意：有一棵点数为 $N$ 的树，以点 1 为根。然后有 $M$ 个操作。

1. 把 $x$ 的点权增加 $a$ 。

2. 把 $x$ 为根的子树中所有点的点权都增加 $a$ 。

3. 询问 $x$ 到根的路径中所有点的点权和。

题解：树链剖分模板题

卡点：我真的好久没打树剖了。。。

1.线段树$tag$写错

2.树剖$query$打错

3.一堆东西没开$long \; long$

 

C++ Code：

#include <cstdio>
#define maxn 100010
using namespace std;
int n, m, S[maxn], s[maxn];
int head[maxn], cnt;
struct Edge {
	int to, nxt;
} e[maxn << 1];
void add(int a, int b) {
	e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
}
int dfn[maxn], son[maxn], sz[maxn], idx;
int dep[maxn], fa[maxn], top[maxn];
void dfs1(int rt) {
	sz[rt] = 1;
	for (int i = head[rt]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (v != fa[rt]) {
			dep[v] = dep[rt] + 1;
			fa[v] = rt;
			dfs1(v);
			if (!son[rt] || sz[v] > sz[son[rt]]) son[rt] = v;
			sz[rt] += sz[v];
		}
	}
}
void dfs2(int rt) {
	dfn[rt] = ++idx;
	int v = son[rt];
	if (v) top[v] = top[rt], dfs2(v);
	for (int i = head[rt]; i; i = e[i].nxt) {
		v = e[i].to;
		if (v != fa[rt] && v != son[rt]) {
			top[v] = v;
			dfs2(v);
		}
	}
}
long long v[maxn << 2], cov[maxn << 2];
void build(int rt, int l, int r) {
	if (l == r) {
		v[rt] = s[l];
		return ;
	}
	int mid = l + r >> 1;
	build(rt << 1, l, mid);
	build(rt << 1 | 1, mid + 1, r);
	v[rt] = v[rt << 1] + v[rt << 1 | 1];
}
void pushdown(int rt, int len) {
	long long &tmp = cov[rt];
	v[rt << 1] += tmp * (len + 1 >> 1);
	v[rt << 1 | 1] += tmp * (len >> 1);
	cov[rt << 1] += tmp;
	cov[rt << 1 | 1] += tmp;
	tmp = 0;
}
void add(int rt, int l, int r, int L, int R, long long num) {
	if (L <= l && R >= r) {
		v[rt] += num * (r - l + 1);
		cov[rt] += num;
		return ;
	}
	int mid = l + r >> 1;
	if (cov[rt]) pushdown(rt, r - l + 1);
	if (L <= mid) add(rt << 1, l, mid, L, R, num);
	if (R > mid) add(rt << 1 | 1, mid + 1, r, L, R, num);
	v[rt] = v[rt << 1] + v[rt << 1 | 1];
}
long long ask(int rt, int l, int r, int L, int R) {
	if (L <= l && R >= r) {
		return v[rt];
	}
	int mid = l + r >> 1;
	long long ans = 0;
//	printf("%d %d %d %d %d\n", rt, l, r, L, R);
	if (cov[rt]) pushdown(rt, r - l + 1);
	if (L <= mid) ans = ask(rt << 1, l, mid, L, R);
	if (R > mid) ans += ask(rt << 1 | 1, mid + 1, r, L, R);
	return ans;
}
long long query(int rt) {
	long long ans = 0;
	while (top[rt] != 1) {
		ans += ask(1, 1, n, dfn[top[rt]], dfn[rt]);
		rt = fa[top[rt]];
	}
	ans += ask(1, 1, n, dfn[1], dfn[rt]);
	return ans;
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) scanf("%d", &S[i]);
	for (int i = 1; i < n; i++) {
		int a, b;
		scanf("%d%d", &a, &b);
		add(a, b);
		add(b, a);
	}
	dep[top[1] = 1] = 1;
	dfs1(1);
	dfs2(1);
	for (int i = 1; i <= n; i++) s[dfn[i]] = S[i];
	build(1, 1, n);
	int op, a; long long x;
	while (m--) {
		scanf("%d%d", &op, &a);
		if (op == 3) {
			printf("%lld\n", query(a));
		} else {
			scanf("%lld", &x);
			if (op == 1) add(1, 1, n, dfn[a], dfn[a], x);
			else add(1, 1, n, dfn[a], dfn[a] + sz[a] - 1, x);
		}
	}
	return 0;
}