神奇的树上启发式合并 (dsu on tree)

参考资料

https://www.cnblogs.com/zhoushuyu/p/9069164.html

https://www.cnblogs.com/candy99/p/dsuontree.html

https://www.cnblogs.com/zcysky/p/6822395.html

简介

树上启发式合并

用到了heavy−light decomposition树链剖分

把轻边子树的信息合并到重链上的点里

因为每次都是先dfs轻儿子再dfs重儿子，只有重儿子子树的贡献保留，所以可以保证dfs到每颗子树时当前全局维护的信息不会有别的子树里的，和莫队很像

算法实现

1.遍历轻儿子

2.遍历重儿子(保留数据)

3.计算所有轻儿子为根的子树

4.如果当前点是轻儿子,清空影响

复杂度分析

树链剖分后每个点到根的路径上有\(logn\)条轻边和\(logn\)条重链

每个点遇见轻边时合并一次,所以至多\(logn\)次

总复杂度\(O(nlogn)\)

例题

CF600E. Lomsat gelral

http://codeforces.com/contest/600/problem/E

题意：询问每颗子树中出现次数最多的颜色们编号和

板子题

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
	x = 0; RG char c = getchar(); bool f = 0;
	while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
	while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
	x = f ? -x : x;
	return ;
}
template<class T> inline void write(T x) {
	if (!x) {putchar(48);return ;}
	if (x < 0) x = -x, putchar('-');
	int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
	for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}
const int N = 1e5+10;
int c[N], n;
struct node {
	int to, nxt;
}g[N<<1];
int last[N], gl;

void add(int x, int y) {
	g[++gl] = (node) {y, last[x]};
	last[x] = gl;
}
int son[N], siz[N];
void dfs1(int u, int f) {
	siz[u] = 1;
	for (int i = last[u]; i; i = g[i].nxt) {
		int v = g[i].to;
		if (v == f) continue;
		dfs1(v, u);
		siz[u] += siz[v];
		if (siz[son[u]] < siz[v]) son[u] = v;
	}
	return ;
}

int num[N], top;
LL sum[N], ans[N];
bool vis[N];

void calc(int u, int fa, int k) {
	sum[num[c[u]]] -= c[u];
	num[c[u]] += k;
	sum[num[c[u]]] += c[u];
	if (sum[top + 1]) top++;
	else if (!sum[top]) top--;

	for (int i = last[u]; i; i = g[i].nxt) {
		int v = g[i].to;
		if (v == fa || vis[v]) continue;
		calc(v, u, k);
	}
	return ;
}

void dfs(int u, int fa, int op) {
	for (int i = last[u]; i; i = g[i].nxt)
		if (g[i].to != fa && g[i].to != son[u])
			dfs(g[i].to, u, 0);
	if (son[u])
		dfs(son[u], u, 1), vis[son[u]] = 1;
	calc(u, fa, 1); vis[son[u]] = 0;
	ans[u] = sum[top];
	if (!op) calc(u, fa, -1);
	return ;
}

int main() {
	read(n);
	for (int i = 1; i <= n; i++) read(c[i]);
	for (int i = 1; i < n; i++) {
		int x, y; read(x), read(y);
		add(x, y), add(y, x);
	}
	dfs1(1, 0);
	dfs(1, 0, 1);
	for (int i = 1; i <= n; i++)
		printf("%I64d ", ans[i]);
	return 0;
}

CF570D Tree Requests

http://codeforces.com/problemset/problem/570/D

https://www.luogu.org/problemnew/show/CF570D

构成回文串,奇数个的字母至多一个

用二进制状压判断即可

\(sum[x]\)表示深度为\(x\)构成的状态

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
	x = 0; RG char c = getchar(); bool f = 0;
	while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
	while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
	x = f ? -x : x;
	return ;
}
template<class T> inline void write(T x) {
	if (!x) {putchar(48);return ;}
	if (x < 0) x = -x, putchar('-');
	int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
	for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}

const int N = 500010;

struct node {
	int to, nxt;
}g[N<<1], q[N];
int last[N], gl;
int n, m;
void add(int x, int y) {
	g[++gl] = (node) {y, last[x]};
	last[x] = gl;
	g[++gl] = (node) {x, last[y]};
	last[y] = gl;
}

char s[N];
int siz[N], son[N], val[N], dep[N], sum[N];
bool vis[N];
void dfs1(int u, int fa) {
	siz[u] = 1;
	for (int i = last[u]; i; i = g[i].nxt) {
		int v = g[i].to;
		if (v == fa) continue;
		dep[v] = dep[u] + 1;
		dfs1(v, u);
		siz[u] += siz[v];
		if (siz[son[u]] < siz[v]) son[u] = v;
	}
}

void calc(int u, int fa) {
	sum[dep[u]] ^= val[u];
	for (int i = last[u]; i; i = g[i].nxt) {
		int v = g[i].to;
		if (v == fa || vis[v]) continue;
		calc(v, u);
	}
	return ;
}

struct Node {
	int h, nxt;
}a[N];
bool ans[N];
int S[N];

bool cal(int x) {
	int tmp = 0;
	while (x) {
		tmp++;
		x -= (x & (-x));
	}
	return tmp <= 1;
}

void dfs(int u, int fa, int op) {
	for (int i = last[u]; i; i = g[i].nxt) {
		int v = g[i].to;
		if (v == fa || son[u] == v) continue;
		dfs(v, u, 0);
	}
	if (son[u]) dfs(son[u], u, 1), vis[son[u]] = 1;
	calc(u, fa); vis[son[u]] = 0;
	for (int i = S[u]; i; i = a[i].nxt)
		ans[i] = cal(sum[a[i].h]);
	if (!op) calc(u, fa);
	return ;
}

int main() {
	read(n), read(m);
	for (int i = 2; i <= n; i++) {
		int x; read(x);
		add(x, i);
	}
	scanf("%s", s+1);
	for (int i = 1; i <= n; i++) val[i] = 1<<(s[i]-'a');
	dep[1] = 1;
	dfs1(1, 0);
	for (int i = 1; i <= m; i++) {
		int h, v;
		read(v), read(h);
		a[i].nxt = S[v];
		S[v] = i; a[i].h = h;
	}
	dfs(1, 0, 1);
	for (int i = 1; i <= m; i++) puts(ans[i] ? "Yes" : "No");
	return 0;
}