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;
}