[Codeforces 487E]Tourists

Description

题库链接

给你张 \(n\) 个点 \(m\) 条边的连通无向图。 \(q\) 次操作，让你支持：

询问可以出现在两点之间的简单路路径上的点的最小权值；修改某个点的点权。

\(1\leq n, m, q\leq 10^5\)

Solution

显然在路径上如果存在一个点双，那么这个点双上的点是都能被经过的。

显然我们可以构建圆方树，方点维护点双中所有的点的点权，用堆实现。

值得注意的是，由于可能一个圆点周围有许多方点，如果更新时在每个方点内都更新的话，复杂度还是不正确的。所以我们规定每个方点只管辖其儿子的圆点。在求 \(lca\) 处特判一下即可。

Code

//It is made by Awson on .3.17#include <bits/stdc++.h>#define LL long long#define dob complex<double>#define Abs(a) ((a) < 0 ? (-(a)) : (a))#define Max(a, b) ((a) > (b) ? (a) : (b))#define Min(a, b) ((a) < (b) ? (a) : (b))#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))#define writeln(x) (write(x), putchar('\n'))#define lowbit(x) ((x)&(-(x)))using namespace std;const int N = 1e5, INF = ~0u>>1;void read(int &x) {char ch; bool flag = 0;for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());x *= 1-2*flag;}void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }int X, n, m, q, u, v, a[(N<<1)+5]; char ch[3];struct Edge {int u, v; };struct Heap {priority_queue<int, vector<int>, greater<int> >p, q;void erase(int o) {q.push(o); }void push(int o) {p.push(o); }int top() {while (!q.empty() && p.top() == q.top()) p.pop(), q.pop(); return p.top(); }}Q[(N<<1)+5];stack<Edge>S;int dfn[N+5], low[N+5], sccno[N+5], times, sccnum, pos;int fa[(N<<1)+5], size[(N<<1)+5], dep[(N<<1)+5], top[(N<<1)+5], son[(N<<1)+5], idx[(N<<1)+5], b[(N<<1)+5];struct graph {struct tt {int to, next; }edge[(N<<2)+5];int path[(N<<1)+5], top;void add(int u, int v) {edge[++top].to = v, edge[top].next = path[u], path[u] = top; }}g1, g2;struct Segment_tree {#define lr(o) (o<<1)#define rr(o) (o<<1|1)int sgm[(N<<3)+5];void build (int o, int l, int r) {if (l == r) {sgm[o] = a[b[l]]; return; } int mid = (l+r)>>1;build(lr(o), l, mid), build(rr(o), mid+1, r);sgm[o] = Min(sgm[lr(o)], sgm[rr(o)]);}int query(int o, int l, int r, int a, int b) {if (a <= l && r <= b) return sgm[o]; int mid = (l+r)>>1;int c1 = INF, c2 = INF;if (a <= mid) c1 = query(lr(o), l, mid, a, b);if (b > mid) c2 = query(rr(o), mid+1, r, a, b);return Min(c1, c2);}void insert(int o, int l, int r, int loc) {if (l == r) {sgm[o] = a[b[l]]; return; } int mid = (l+r)>>1;if (loc <= mid) insert(lr(o), l, mid, loc); else insert(rr(o), mid+1, r, loc);sgm[o] = Min(sgm[lr(o)], sgm[rr(o)]);}}T;void tarjan(int o, int fa) {dfn[o] = low[o] = ++times;for (int i = g1.path[o]; i; i = g1.edge[i].next) {int v = g1.edge[i].to;if (!dfn[v]) {S.push((Edge){o, v});tarjan(v, o); low[o] = Min(low[o], low[v]);if (low[v] >= dfn[o]) {++n; ++sccnum;while (true) {Edge e = S.top(); S.pop();if (sccno[e.u] != sccnum) sccno[e.u] = sccnum, g2.add(n, e.u), g2.add(e.u, n);if (sccno[e.v] != sccnum) sccno[e.v] = sccnum, g2.add(n, e.v), g2.add(e.v, n);if (e.u == o && e.v == v) break;}}}else if (v != fa) low[o] = Min(low[o], dfn[v]);}}void dfs1(int o, int depth, int father) {fa[o] = father, dep[o] = depth, size[o] = 1;for (int i = g2.path[o]; i; i = g2.edge[i].next)if (g2.edge[i].to != father) {if (o >= X) Q[o].push(a[g2.edge[i].to]);dfs1(g2.edge[i].to, depth+1, o);size[o] += size[g2.edge[i].to];if (size[g2.edge[i].to] > size[son[o]]) son[o] = g2.edge[i].to;}if (o > X) a[o] = Q[o].top();}void dfs2(int o, int tp) {top[o] = tp, idx[o] = ++pos, b[pos] = o;if (son[o]) dfs2(son[o], tp);for (int i = g2.path[o]; i; i = g2.edge[i].next)if (g2.edge[i].to != fa[o] && g2.edge[i].to != son[o])dfs2(g2.edge[i].to, g2.edge[i].to);}int query(int u, int v) {int ans = INF, t;while (top[u] != top[v]) {if (dep[top[u]] < dep[top[v]]) Swap(u, v);t = T.query(1, 1, n, idx[top[u]], idx[u]);ans = Min(ans, t); u = fa[top[u]];}if (dep[u] < dep[v]) Swap(u, v);t = T.query(1, 1, n, idx[v], idx[u]); ans = Min(ans, t);if (v > X) t = T.query(1, 1, n, idx[fa[v]], idx[fa[v]]), ans = Min(ans, t);return ans;}void update(int o, int val) {if (fa[o]) Q[fa[o]].erase(a[o]);a[o] = val; T.insert(1, 1, n, idx[o]);if (fa[o]) Q[fa[o]].push(a[o]), a[fa[o]] = Q[fa[o]].top(), T.insert(1, 1, n, idx[fa[o]]);}void work() {read(n), read(m), read(q); X = n;for (int i = 1; i <= n; i++) read(a[i]);for (int i = 1; i <= m; i++) read(u), read(v), g1.add(u, v), g1.add(v, u);tarjan(1, 0); dfs1(1, 1, 0); dfs2(1, 1); T.build(1, 1, n);while (q--) {scanf("%s", ch); read(u), read(v);if (ch[0] == 'A') writeln(query(u, v));else update(u, v);}}int main() {work(); return 0;}