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Query on a tree V VI VII [LCT]

青鸾惊鸿 2022-07-12 阅读 91
#definegiti++OpenStack云计算

QTREE5

操作 : 反转某个点的颜色, 查询某个点的最近白点

我们维护Splay中, 最浅节点到最近白点的距离(lsum), 以及最深节点到它的最近白点的距离(rsum)

那么查询x的答案很明显 Accss(x), Splay(x), x -> rsum 就是答案

考虑如何 Pushup

Query on a tree V VI VII [LCT]_i++

Minson是虚子树答案, 用一个set维护

#include<bits/stdc++.h>
#define N 100050
using namespace std;
int read(){
  int cnt = 0; char ch = 0;
  while(!isdigit(ch)) ch = getchar();
  while(isdigit(ch)) cnt = cnt*10 + (ch-'0'), ch = getchar();
  return cnt;
}
const int inf = 0x3fffffff;
vector<int> v[N];
int n, m, col[N];
int ch[N][2], fa[N], siz[N], lsum[N], rsum[N];
multiset <int> S[N]; 
void dfs(int u, int f){
  for(int i=0; i<v[u].size(); i++){
    int t = v[u][i]; if(t == f) continue;
    fa[t] = u; S[u].insert(inf + 1); dfs(t, u);
  } 
}
#define ls ch[x][0]
#define rs ch[x][1]
bool isRoot(int x){ return ch[fa[x]][0] != x && ch[fa[x]][1] != x;}
int Top(int x){ if(S[x].empty()) return inf; return *S[x].begin();}
void Pushup(int x){
  if(!x) return; lsum[x] = rsum[x] = inf;
  siz[x] = siz[ls] + siz[rs] + 1;
  lsum[x] = min(lsum[ls], siz[ls] + min(col[x] ? 0 : inf, min(Top(x), lsum[rs] + 1)));
  rsum[x] = min(rsum[rs], siz[rs] + min(col[x] ? 0 : inf, min(Top(x), rsum[ls] + 1)));
}
void rotate(int x){
  int y = fa[x], z = fa[y], k = ch[y][1] == x;
  if(!isRoot(y)) ch[z][ch[z][1] == y] = x; fa[x] = z;
  ch[y][k] = ch[x][k^1]; fa[ch[x][k^1]] = y;
  ch[x][k^1] = y; fa[y] = x; Pushup(y); Pushup(x);
}
void Splay(int x){
  while(!isRoot(x)){
    int y = fa[x], z = fa[y];
    if(!isRoot(y)) rotate((ch[y][0] == x) ^ (ch[z][0] == y) ? x : y);
    rotate(x);
  }
}
void Access(int x){
  for(int y = 0; x; y = x, x = fa[x]){
    Splay(x); 
    if(rs) S[x].insert(lsum[rs] + 1);
    rs = y;
    if(rs) S[x].erase(S[x].find(lsum[rs] + 1));
    Pushup(x);
  }
}
int main(){
  n = read(); lsum[0] = rsum[0] = inf;
  for(int i=1; i<n; i++){
    int x = read(), y = read(); 
    v[x].push_back(y); v[y].push_back(x); 
  } dfs(1, 0);
  m = read();
  while(m--){
    int op = read(), x = read();
    if(op == 0){ Access(x); Splay(x); col[x] ^= 1; Pushup(x);}
    if(op == 1){ Access(x); Splay(x); printf("%d\n", (rsum[x] == inf) ? -1 : rsum[x]);}
  } return 0;
}

QTREE6 

操作 : 反转颜色, 查询同色连通块的大小

首先开两颗LCT, 反转颜色相当于在一边暴力 Cut, 一边暴力Link

我们考虑只Cut x与它的fa, 也之Link x与它的fa, 这样一个连通块只有可能 rt与其颜色不同, 特判一下就好

然后就是维护虚子树的Siz

#include<bits/stdc++.h>
#define N 1000050
using namespace std;
int read(){
  int cnt = 0; char ch = 0;
  while(!isdigit(ch)) ch = getchar();
  while(isdigit(ch)) cnt = cnt*10 + (ch-'0'), ch = getchar();
  return cnt;
}
int first[N], nxt[N], to[N], tot;
void add(int x, int y){ nxt[++tot] = first[x], first[x] = tot, to[tot] = y;}
int n, m, col[N];
struct LCT{
  struct Node{ int ch[2], fa, siz; } t[N];
  int si[N];
  #define ls t[x].ch[0] 
  #define rs t[x].ch[1]
  bool isRoot(int x){ int fa = t[x].fa; return t[fa].ch[0] != x && t[fa].ch[1] != x; }
  void Pushup(int x){ t[x].siz = t[ls].siz + t[rs].siz + si[x] + 1;}
  void rotate(int x){
    int y = t[x].fa, z = t[y].fa, k = t[y].ch[1] == x;
    if(!isRoot(y)) t[z].ch[t[z].ch[1] == y] = x; t[x].fa = z;
    t[y].ch[k] = t[x].ch[k^1]; t[t[x].ch[k^1]].fa = y;
    t[x].ch[k^1] = y; t[y].fa = x; Pushup(y); Pushup(x);
  }
  void Splay(int x){
    while(!isRoot(x)){
      int y = t[x].fa, z = t[y].fa;
      if(!isRoot(y)) rotate((t[y].ch[0] == x) ^ (t[z].ch[0] == y) ? x : y);
      rotate(x);
    } Pushup(x);
  }
  void Access(int x){
    for(int y=0; x; y=x, x=t[x].fa){ 
      Splay(x);
      si[x] += t[rs].siz;
      rs = y;
      si[x] -= t[rs].siz;
      Pushup(x);
    } 
  }
  void Link(int x, int y){ 
    if(!y) return;
    Access(y); Splay(y); Splay(x); t[x].fa = y; si[y] += t[x].siz; Pushup(y); 
  }
  void Cut(int x, int y){
    if(!y) return;
    Access(x); Splay(x); t[ls].fa = 0; ls = 0; Pushup(x);
  }
  int Findroot(int x){
    Access(x); Splay(x);
    while(ls) x = ls; Splay(x);
    return x; 
  }
  int Quary(int x){
    int c = col[x]; x = Findroot(x);
    return col[x] == c ? t[x].siz : t[rs].siz;
  }
}lct[2]; 
int fa[N];
void dfs(int u, int f){
  for(int i=first[u];i;i=nxt[i]){
    int t = to[i]; if(t == f) continue;
    fa[t] = u; lct[0].Link(t, u); dfs(t, u);
  }
}
int main(){
  n = read(); 
  for(int i=1; i<n; i++){
    int x = read(), y = read();
    add(x, y); add(y, x);
  } dfs(1, 0); m = read();
  while(m--){
    int op = read(), x = read();
    if(op == 0) printf("%d\n", lct[col[x]].Quary(x));
    if(op == 1){ lct[col[x]].Cut(x, fa[x]); col[x] ^= 1; lct[col[x]].Link(x, fa[x]);}
  } return 0;
}

QTREE7 

与上一道题类似, 维护同色连通块的最大值, multiset就可以了

#include<bits/stdc++.h>
#define N 200050
using namespace std;
int read(){
  int cnt = 0, f = 1; char ch = 0;
  while(!isdigit(ch)){ ch = getchar(); if(ch == '-') f = -1;}
  while(isdigit(ch)) cnt = cnt*10 + (ch-'0'), ch = getchar();
  return cnt * f;
}
int first[N], nxt[N], to[N], tot;
void add(int x, int y){nxt[++tot] = first[x], first[x] = tot, to[tot] = y;}
int n, m, col[N], f[N], w[N];
struct Node{
  int ch[N][2], fa[N], Max[N];
  multiset<int> S[N];
  bool isRoot(int x){ return ch[fa[x]][0] != x && ch[fa[x]][1] != x;}
  void Pushup(int x){
    Max[x] = max(w[x], max(Max[ch[x][0]], Max[ch[x][1]]));
    if(S[x].size()) Max[x] = max(Max[x], *S[x].rbegin());
  }
  void rotate(int x){
    int y = fa[x], z = fa[y], k = ch[y][1] == x;
    if(!isRoot(y)) ch[z][ch[z][1] == y] = x; fa[x] = z;
    ch[y][k] = ch[x][k^1]; fa[ch[x][k^1]] = y;
    ch[x][k^1] = y; fa[y] = x; Pushup(y); Pushup(x);
  }
  void Splay(int x){
    while(!isRoot(x)){
      int y = fa[x], z = fa[y];
      if(!isRoot(y)) rotate((ch[y][1]==x) ^ (ch[z][1]==y) ? x : y);
      rotate(x); 
    } Pushup(x);
  }
  void Access(int x){
    for(int y = 0; x; y = x, x = fa[x]){
      Splay(x); 
      if(ch[x][1]) S[x].insert(Max[ch[x][1]]);
      ch[x][1] = y; 
      if(ch[x][1]) S[x].erase(Max[ch[x][1]]);
      Pushup(x);
    }
  }
  int Findroot(int x){
    Access(x); Splay(x); while(ch[x][0]) x = ch[x][0]; Splay(x); return x;
  }
  void Link(int x, int y){
    if(!y) return;
    Access(y); Splay(y); Splay(x);
    fa[x] = y; S[y].insert(Max[x]); Pushup(y);
  }
  void Cut(int x, int y){
    if(!y) return;
    Access(x); Splay(x); fa[ch[x][0]] = 0; ch[x][0] = 0; Pushup(x);
  }
  int Quary(int x){
    int c = col[x]; x = Findroot(x);
    return col[x] == c ? Max[x] : Max[ch[x][1]];
  }
  void Modify(int x){
    Access(x); Splay(x); w[x] = read(); Pushup(x);
  }
}T[2];
void dfs(int u, int ff){
  for(int i=first[u];i;i=nxt[i]){
    int t = to[i]; if(t == ff) continue;
    T[col[t]].Link(t, u); f[t] = u; dfs(t, u); 
  }
}
int main(){
  n = read();
  for(int i=1; i<n; i++){
    int x = read(), y = read();
    add(x, y); add(y, x);
  } 
  for(int i=1; i<=n; i++) col[i] = read();
  for(int i=1; i<=n; i++) w[i] = read();
  T[0].Max[0] = T[1].Max[0] = -2e9;
  dfs(1, 0); m = read();
  while(m--){
    int op = read(), u = read();
    if(op == 0) printf("%d\n", T[col[u]].Quary(u));
    if(op == 1){ T[col[u]].Cut(u, f[u]); col[u] ^= 1; T[col[u]].Link(u, f[u]);}
    if(op == 2){ T[col[u]].Modify(u);}
  } return 0;
}

 


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