[SDOI2013]森林

标签：lib   define   pre   ++i   就会   mat   节点   回收   ios   

嘟嘟嘟


如果不是森林，很自然想到树上主席树。
但现在他是个森林，那就lct+主席树？
不，怎么会那么毒瘤。
只要启发式合并就好啦。


不用垃圾回收（主席树启发式合并垃圾回收似乎很麻烦）。需要注意的是，一定要处理好较小树节点的新的元素，包括深度，父亲……最重要的是求lca的倍增数组，一定要都清空，因为合并到新的树后他的深度可能没有原来那么深，所以更远的祖先就不存在了。但是如果不清空的话，倍增的时候就会跳到那里。毕竟此时fa[u][x] ^ fa[i][y]是成立的。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 8e4 + 5;
const int N = 18;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

char s[2];
int ID, _n, n, m, q;
int a[maxn], b[maxn];
struct Edge
{
  int nxt, to;
}e[maxn << 1];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y)
{
  e[++ecnt] = (Edge){head[x], y};
  head[x] = ecnt;
}

struct Tree
{
  int ls, rs, sum;
}t[maxn * 400];
int root[maxn], tcnt = 0;
In void insert(int old, int& now, int l, int r, int id)
{
  t[now = ++tcnt] = t[old];
  ++t[now].sum;
  if(l == r) return;
  int mid = (l + r) >> 1;
  if(id <= mid) insert(t[old].ls, t[now].ls, l, mid, id);
  else insert(t[old].rs, t[now].rs, mid + 1, r, id);
}
In int query(int pre, int Lca, int x, int y, int l, int r, int k)
{
  if(l == r) return l;
  int mid = (l + r) >> 1, Sum = t[t[x].ls].sum + t[t[y].ls].sum - t[t[pre].ls].sum - t[t[Lca].ls].sum;
  if(k <= Sum) return query(t[pre].ls, t[Lca].ls, t[x].ls, t[y].ls, l, mid, k);
  else return query(t[pre].rs, t[Lca].rs, t[x].rs, t[y].rs, mid + 1, r, k - Sum);
}

int siz[maxn], bel[maxn];
int dep[maxn], fa[N + 2][maxn];
In int lca(int x, int y)
{
  if(dep[x] < dep[y]) swap(x, y);
  for(int i = N; i >= 0; --i)
    if(dep[x] - (1 << i) >= dep[y]) x = fa[i][x];
  if(x == y) return x;
  for(int i = N; i >= 0; --i)
    if(fa[i][x] ^ fa[i][y]) x = fa[i][x], y = fa[i][y];
  return fa[0][x];
}
In void dfs(int now, int _f, int rot)
{
  siz[now] = 1; bel[now] = rot;
  for(int i = 1; i <= N; ++i) fa[i][now] = 0;
  for(int i = 1; (1 << i) <= dep[now]; ++i)
    fa[i][now] = fa[i - 1][fa[i - 1][now]];
  insert(root[_f], root[now], 1, _n, a[now]);
  for(int i = head[now], v; ~i; i = e[i].nxt)
    {
      if((v = e[i].to) == _f) continue;
      fa[0][v] = now, dep[v] = dep[now] + 1;
      dfs(v, now, rot);
      siz[now] += siz[v];
    }
}

int main()
{
  Mem(head, -1);
  ID = read();
  n = read(), m = read(), q = read();
  for(int i = 1; i <= n; ++i) a[i] = b[i] = read();
  sort(b + 1, b + n + 1);
  _n = unique(b + 1, b + n + 1) - b - 1;
  for(int i = 1; i <= n; ++i)
    a[i] = lower_bound(b + 1, b + _n + 1, a[i]) - b;
  for(int i = 1; i <= m; ++i)
    {
      int x = read(), y = read();
      addEdge(x, y), addEdge(y, x);
    }
  for(int i = 1; i <= n; ++i) if(!siz[i]) dfs(i, 0, i);
  for(int i = 1, ans = 0; i <= q; ++i)
    {
      scanf("%s", s); int x = read() ^ ans, y = read() ^ ans;
      if(s[0] == 'Q')
    {
      int K = read() ^ ans, z = lca(x, y);
      ans = b[query(root[fa[0][z]], root[z], root[x], root[y], 1, _n, K)];
      write(ans), enter;
    }
      else
    {
      addEdge(x, y), addEdge(y, x);
      if(siz[bel[x]] > siz[bel[y]]) swap(x, y);
      fa[0][x] = y, dep[x] = dep[y] + 1, siz[bel[y]] += siz[bel[x]]; 
      dfs(x, y, bel[y]);
    }
    }
  return 0;
}

[SDOI2013]森林

标签：lib   define   pre   ++i   就会   mat   节点   回收   ios   

原文地址：https://www.cnblogs.com/mrclr/p/10645063.html