【洛谷P4172】水管局长

题目大意：给定 N 个点，M 条边的无向图，支持两种操作：动态删边和查询任意两点之间路径上边权的最大值最小是多少。

题解：
引理：对原图求最小生成树，可以保证任意两点之间的路径上边权的最大值取得最小值。
证明：任取两点 x, y，若 x, y 的路径上最大值最小的边不在最小生成树的路径上，可以将那条边加入最小生成树中，并删去由这条边的加入所带来的环中边权最大的那条边，可以使得最小生成树更小，产生矛盾，证毕。
有了引理之后，问题转化成了维护支持动态删边的最小生成树。发现删边可能会导致最少生成树不断发生变化，每次变化都需要重构生成树，时间复杂度较高。可以采用离线询问，转化成倒序加边的最小生成树的维护，每次加一条边时，只需删除环上最大的那条边即可。支持动态加边和删边，可以采用 lct 进行维护。最后，可以进行点边转化，即：将边缩为一个点，边权为点权，点的点权为 0 即可。

代码如下

#include <bits/stdc++.h>

using namespace std;

struct edge {
	int x, y, z;
	bool has;
	bool operator<(const edge &rhs) {
		return this->z < rhs.z;
	}
};

struct UFS {
	vector<int> f;
	UFS(int n) {
		f.resize(n + 1);
		for (int i = 1; i <= n; i++) {
			f[i] = i;
		}
	}
	int find(int x) {
		return x == f[x] ? x : f[x] = find(f[x]);
	}
	bool merge(int x, int y) {
		x = find(x), y = find(y);
		if (x != y) {
			f[x] = y;
			return 1;
		}
		return 0;
	}
};

struct node {
	node *l, *r, *p;
	int val, maxv, rev, id;
	node (int _val = 0, int _id = 0) {
		l = r = p = NULL;
		val = maxv = _val;
		id = _id;
		rev = 0;
	}
	void unsafe_reverse() {
		swap(l, r);
		rev ^= 1;
	}
	void pull() {
		maxv = val;
		if (l != NULL) {
			l->p = this;
			maxv = max(maxv, l->maxv);
		} 
		if (r != NULL) {
			r->p = this;
			maxv = max(maxv, r->maxv);
		}
	}
	void push() {
		if (rev) {
			if (l != NULL) {
				l->unsafe_reverse();
			}
			if (r != NULL) {
				r->unsafe_reverse();
			}
			rev = 0;
		}
	}
};
bool is_root(node *v) {
	if (v == NULL) {
		return false;
	}
	return (v->p == NULL) || (v->p->l != v && v->p->r != v);
}
void rotate(node *v) {
  node *u = v->p;
  assert(u != NULL);
  v->p = u->p;
  if (v->p != NULL) { // work with father
    if (u == v->p->l) {
      v->p->l = v;
    }
    if (u == v->p->r) {
      v->p->r = v;
    }
  }
  if (v == u->l) {
    u->l = v->r;
    v->r = u;
  }
  if (v == u->r) {
    u->r = v->l;
    v->l = u;
  }
  u->pull();
  v->pull();
}
void deal_with_push(node *v) {
  static stack<node*> stk;
  while (true) {
    stk.push(v);
    if (is_root(v)) {
      break;
    }
    v = v->p;
  }
  while (!stk.empty()) {
    stk.top()->push();
    stk.pop();
  }
}
void splay(node *v) {
  deal_with_push(v);
  while (!is_root(v)) {
    node *u = v->p;
    if (!is_root(u)) {
      if ((u->p->l == u) ^ (u->l == v)) {
        rotate(v);
      } else {
        rotate(u);
      }
    }
    rotate(v);
  }
}
void access(node *v) {
  node *u = NULL;
  while (v != NULL) {
    splay(v);
    v->r = u;
    v->pull();
    u = v;
    v = v->p;
  }
}
void make_root(node *v) {
  access(v);
  splay(v);
  v->unsafe_reverse();
}
node *find_root(node *v) {
  access(v);
  splay(v);
  while (v->l != NULL) {
    v->push();
    v = v->l;
  }
  splay(v);
  return v;
}
void split(node *u, node *v) {
  make_root(u);
  access(v);
  splay(v);
}
void link(node *u, node *v) {
  if (find_root(u) == find_root(v)) {
    return;
  }
  make_root(v);
  v->p = u;
}
void cut(node *u, node *v) {
  make_root(u);
  if (find_root(v) == u && v->p == u && v->l == NULL) {
    v->p = u->r = NULL;
    u->pull();
  }
}
node* find(node *v, int val) {
	while (true) {
		if (v->val == val) {
			break;
		}
		if (v->l != NULL && v->l->maxv == val) {
			v = v->l;
		} else {
			v = v->r;
		}
	}
	return v;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int n, m, q;
	cin >> n >> m >> q;
	vector<node*> t(n + m + 1);
	for (int i = 1; i <= n; i++) {
		t[i] = new node(0, i);
	}
	vector<edge> e(m + 1);
	map<pair<int, int>, int> id;
	for (int i = 1; i <= m; i++) {
		int x, y, z;
		cin >> x >> y >> z;
		if (x > y) {
			swap(x, y);
		}
		e[i] = {x, y, z, 1};
		id[{x, y}] = i;
	}
	vector<pair<int, pair<int, int>>> events(q + 1);
	for (int i = 1; i <= q; i++) {
		int opt, x, y;
		cin >> opt >> x >> y;
		if (x > y) {
			swap(x, y);
		}
		events[i] = {opt, {x, y}};
		if (opt == 2) {
			e[id[{x, y}]].has = 0;
		}
	}
	auto kruskal = [&](vector<edge> ee) {
		sort(ee.begin() + 1, ee.end());
		UFS ufs(n);
		for (int i = 1; i <= m; i++) {
			if (ee[i].has == 1) {
				int x = ee[i].x, y = ee[i].y, z = ee[i].z;
				if (ufs.merge(x, y) == 1) {
					int _id = id[{x, y}] + n; // edge_id
					t[_id] = new node(z, _id);
					link(t[x], t[_id]);
					link(t[y], t[_id]);
				}
			}
		}
	};
	kruskal(e);
	vector<int> ans;
	for (int i = q; i >= 1; i--) {
		int x = events[i].second.first;
		int y = events[i].second.second;
		int eid = id[{x, y}];
		int z = e[eid].z;
		if (events[i].first == 1) {
			split(t[x], t[y]);
			ans.push_back(t[y]->maxv);
		} else {
			split(t[x], t[y]);
			if (t[y]->maxv > z) {
				int _id = find(t[y], t[y]->maxv)->id; // edge id
				int a = e[_id - n].x, b = e[_id - n].y;
				cut(t[a], t[_id]);
				cut(t[b], t[_id]);
				t[eid + n] = new node(z, eid + n);
				link(t[x], t[eid + n]);
				link(t[y], t[eid + n]);
			}
		}
	}
	reverse(ans.begin(), ans.end());
	for (auto v : ans) {
		cout << v << endl;
	}
	return 0;
}