BZOJ 3589 动态树 （树链剖分+线段树）

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前言

众所周知， $90\%$的题目与解法毫无关系。

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题意

有一棵有根树，两种操作。一种是子树内每一个点的权值加上一个同一个数，另一种是查询多条路径的并的点权之和。

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分析

很容易看出是树链剖分+线段树的题目，唯一的问题就是多条路径可能有交集。那么我们只要把每条路径拆成多个部分，每一部分是某重链上连续的一段，就得到了很多区间。然后排序取并集就能在线段树上操作了。

AC CODE

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
const int mod = 2147483647;
int n, m;
struct SegmentTree {
	int v[MAXN<<2], lz[MAXN<<2], l[MAXN<<2], r[MAXN<<2], len[MAXN<<2];
	inline void pushdown(int i) {
		if(lz[i]) {
			v[i<<1] += lz[i]*len[i<<1], lz[i<<1] += lz[i];
			v[i<<1|1] += lz[i]*len[i<<1|1], lz[i<<1|1] += lz[i];
			lz[i] = 0;
		}
	}
	void build(int i, int L, int R) {
		l[i] = L, r[i] = R, len[i] = R-L+1;
		if(L == R) return;
		build(i<<1, L, (L+R)>>1);
		build(i<<1|1, (L+R)/2+1, R);
	}
	void Modify(int i, int L, int R, int k) {
		if(L <= l[i] && r[i] <= R) { v[i] += k*len[i], lz[i] += k; return; }
		pushdown(i);
		int mid = (l[i] + r[i]) >> 1;
		if(L <= mid) Modify(i<<1, L, R, k);
		if(R > mid) Modify(i<<1|1, L, R, k);
		v[i] = v[i<<1] + v[i<<1|1];
	}
	int Query(int i, int L, int R) {
		if(L <= l[i] && r[i] <= R) return v[i];
		pushdown(i);
		int mid = (l[i] + r[i]) >> 1, res = 0;
		if(L <= mid) res += Query(i<<1, L, R);
		if(R > mid) res += Query(i<<1|1, L, R);
		return res;
	}
}T;
int fir[MAXN], to[MAXN<<1], nxt[MAXN<<1], cnt;
inline void add(int x, int y) {
	to[++cnt] = y; nxt[cnt] = fir[x]; fir[x] = cnt;
}
int dep[MAXN], top[MAXN], dfn[MAXN], hson[MAXN], sz[MAXN], fa[MAXN];
inline void read(int &num) {
	char ch; while((ch=getchar())<'0'||ch>'9');
	for(num=0;ch>='0'&&ch<='9';num=num*10+ch-'0',ch=getchar());
}
struct node {
	int l, r;
	inline bool operator <(const node &t)const {
		return l == t.l ? r > t.r : l < t.l;
	}
}a[MAXN], b[MAXN];
int cur;
inline void pre(int x, int y) {
	while(top[x] != top[y]) {
		if(dep[top[x]] < dep[top[y]]) swap(x, y);
		a[++cur] = (node){ dfn[top[x]], dfn[x] };
		x = fa[top[x]];
	}
	if(dep[x] < dep[y]) swap(x, y);
	a[++cur] = (node){ dfn[y], dfn[x] };
}
inline int solve() {
	sort(a + 1, a + cur + 1);
	int tot = 0, res = 0, l = a[1].l, r = a[1].r;
	for(int i = 1; i <= cur; ++i)
		if(a[i].r > r) {
			if(a[i].l > r+1) b[++tot] = (node){ l, r }, l = a[i].l, r = a[i].r;
			else r = a[i].r;
		}
	b[++tot] = (node){ l, r };
	for(int i = 1; i <= tot; ++i)
		res = (res + T.Query(1, b[i].l, b[i].r)) & mod;
	return res;
}

inline void dfs(int u, int ff) {
	dep[u] = dep[fa[u]=ff] + (sz[u]=1);
	for(int i = fir[u]; i; i = nxt[i])
		if(to[i] != fa[u]) {
			dfs(to[i], u), sz[u] += sz[to[i]];
			if(sz[to[i]] > sz[hson[u]]) hson[u] = to[i];
		}
}

inline void dfs2(int u, int tp) {
	top[u] = tp; dfn[u] = ++cur;
	if(hson[u]) dfs2(hson[u], tp);
	for(int i = fir[u]; i; i = nxt[i])
		if(to[i] != fa[u] && to[i] != hson[u])
			dfs2(to[i], to[i]);
}

int main () {
	read(n);
	for(int i = 1, x, y; i < n; ++i)
		read(x), read(y), add(x, y), add(y, x);
	dfs(1, 0); dfs2(1, 1);
	T.build(1, 1, n);
	read(m);
	int opt, x, y;
	while(m--) {
		read(opt);
		if(!opt) read(x), read(y), T.Modify(1, dfn[x], dfn[x]+sz[x]-1, y);
		else {
			read(opt); cur = 0;
			while(opt--) read(x), read(y), pre(x, y);
			printf("%d\n", solve());
		}
	}
}

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$Upd$

不要问我中间为什么不取模。因为懒。能过就行

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$EOF$