As a Review.

Typora even cannot deal with $\LaTeX$ in sudo mode?

Des.

Giving a undirected graph of $N$ nodes and $M$ edges. And you are given $Q$ operations, they might be deleting an edge or ask if there’re at least 2 paths without intersection between 2 nodes.

Sol.

The situation in description is exactly what a biconnected-component of a graph is, as what we called BCC.

However deleting operation is difficult to execute, so we reverse the operation sequence and answer questions off-line.

When edges on the tree form a loop, a BCC is generated, and we shrink all nodes on that loop into a specific node, this can be maintained with a UFS.

When answering, just check if the two nodes are in the same component.

Notice that as we shrink nodes, operation with $f[x]$ in LCT should be modified to $bel[f[i]]$, which is the representing node of one component.

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#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>

using namespace std;
using namespace __gnu_pbds;
typedef gp_hash_table<int, int>::iterator iter;

const int maxn = 1e+5 + 5;
const bool ANSWER_YES = true;
const bool ANSWER_NO = false;

struct querys {
  int typ, u, v;
}q[maxn];

inline int rd() {
  register int x = 0, c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();
  return x;
}
int n, m, Q, ans[maxn];
gp_hash_table<int, int> e_to_id[maxn];
int rev[maxn], ch[maxn][2], f[maxn], bel[maxn], fa[maxn];

int belong(int x) {
  return bel[x] == x ? x : bel[x] = belong(bel[x]);
}
int find(int x) {
  return fa[x] == x ? x : fa[x] = find(fa[x]);
}
inline int getid(int x) {
  return ch[belong(f[x])][1] == x;
}
inline int isroot(int x) {
  return ch[belong(f[x])][0] != x && ch[belong(f[x])][1] != x;
}
inline void pushdown(int x) {
  if (rev[x]) {
    rev[ch[x][0]] ^= 1;
    rev[ch[x][1]] ^= 1;
    swap(ch[x][0], ch[x][1]);
    rev[x] = 0;
  }
}
inline void pushpath(int x) {
  if (!isroot(x)) pushpath(belong(f[x]));
  pushdown(x);
}
inline void rotate(int x) {
  int y = belong(f[x]), z = belong(f[y]), k = getid(x);
  if (!isroot(y)) 
    ch[z][ch[z][1] == y] = x;
  ch[y][k] = ch[x][k ^ 1];
  ch[x][k ^ 1] = y;
  f[y] = x; f[ch[y][k]] = y; f[x] = z;
}
inline void splay(int x) {
  pushpath(x);
  for (; !isroot(x); rotate(x)) {
    int y = belong(f[x]);
    if (!isroot(y))
      rotate(getid(y) == getid(x) ? y : x);
  }
}
inline void access(int x) {
  for (int y = 0; x; y = x, x = belong(f[x]))
    splay(x), ch[x][1] = y;
}
inline void makeroot(int x) {
  access(x); splay(x); rev[x] ^= 1;
}
inline void split(int x, int y) {
  makeroot(x); access(y); splay(y); 
}
inline void link(int x, int y) {
  makeroot(x); f[x] = y;
}
void merge(int x, int y) {
  bel[belong(x)] = y; pushdown(x);
  if (ch[x][0]) merge(ch[x][0], y);
  ch[x][0] = 0;
  if (ch[x][1]) merge(ch[x][1], y);
  ch[x][1] = 0;
}
inline void add(int x, int y) {
  x = belong(x); y = belong(y);
  if (x == y) return;
  int fx = find(x), fy = find(y);
  if (fx != fy) {
    fa[fx] = fy;
    link(x, y);
  } else {
    split(x, y);
    merge(y, y);
  }
}
inline bool query(int x, int y) {
  x = belong(x); y = belong(y);
  if (x == y) return ANSWER_YES;
  else return ANSWER_NO;
}

int main() {
  n = rd(); m = rd(); Q = rd();
  int u, v;
  for (int i = 1; i <= n; ++i)
    fa[i] = bel[i] = i;
  for (int i = 1; i <= m; ++i) {
    u = rd();
    v = rd();
    int mn = min(u, v), mx = max(u, v);
    e_to_id[mn][mx] = i;
  }
  char opt[5];
  for (int i = 1; i <= Q; ++i) {
    scanf("%s", opt);
    if (opt[0] == 'Z') {
      q[i].typ = 0;
      q[i].u = rd(); q[i].v = rd();
      int mn = min(q[i].u, q[i].v), mx = max(q[i].u, q[i].v);
      e_to_id[mn][mx] = 0;
    } else {
      q[i].typ = 1;
      q[i].u = rd(); q[i].v = rd();
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (iter p = e_to_id[i].begin(); p != e_to_id[i].end(); ++p) {
      if (p->second == 0) continue;
      else add(i, p->first);//, printf("%d %d\n", i, p.first);
    }
  }
  for (int i = Q; i; --i) {
    if (q[i].typ == 0) {
      add(q[i].u, q[i].v);
    } else {
      ans[i] = query(q[i].u, q[i].v);
    }
  }
  for (int i = 1; i <= Q; ++i)
    if (q[i].typ)
      puts(ans[i] ? "Yes" : "No");
}