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;
}