CF487E Tourists

题面

题目大意:

给定一张 $ n $ 个点 \(m\) 条边的无向图，每个点都有权值 \(w_i\)，要处理 \(q\)​ 个操作

• 将某个点的权值修改
• 询问两点间路径中点的最小权值

$ 1\leq m,q,n \leq 10^5, 1\leq w_i\leq 10^9$

solution

知识点：圆方树

什么是圆方树

无向图，对于每个点双（任意两点都有两条路径可以到达），建一个方点，让方点和它对应的点双中的圆点连边。

如何构建圆方树？

用 \(tarjan\)​ 求出每个点双，新建一个节点，连边就好了

因为是求路径上的最小值，可以考虑把方点的权值赋为所在点双中所有点权值的最小值，因为每次修改都要比较方点周围的所有点，所以遇见菊花图就很容易被卡成 \(O(nq)\)

因为圆方树是一棵树，所以考虑树的性质，对于每个方点权值赋为它所有儿子权值的最小值，用 \(multiset\) 维护就好了

multiset

code

/*
work by:Ariel_
建立圆方树，方点开 multiset 存周围点的最小值，然后树剖求链上最小值就好了 
*/
#include<iostream>
#include<cstdio>
#include<vector>
#include<set>
#define lson rt << 1
#define rson rt << 1|1
#define ll long long
#define rg register
using namespace std;
const int MAXN = 2e5 + 5;
const int INF = 0x3f3f3f3f;
int read(){
    int x = 0,f = 1; char c = getchar();
    while(c < '0'||c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') {x = x*10 + c - '0'; c = getchar();}
    return x*f;
}
int n, m, q, w[MAXN], fcnt;
int fa[MAXN], dep[MAXN], siz[MAXN], son[MAXN], top[MAXN], id[MAXN], o, val[MAXN];
multiset<int> S[200010];
struct edge {int v, nxt;}e[MAXN << 1], E[MAXN << 1];
int Head[MAXN], head[MAXN], Ecnt, ecnt;
void Add_edge(int u, int v) {
   E[++Ecnt] = (edge){v, Head[u]};
   Head[u] = Ecnt;
} 
void add_edge(int u, int v)  {
   e[++ecnt] = (edge) {v, head[u]};
   head[u] = ecnt;
}
namespace Seg{ 
   struct Tree {
     int l, r, minn;
   }tree[MAXN << 3];
   void push_up(int rt) {
   	  tree[rt].minn = min(tree[lson].minn, tree[rson].minn);
   }
   void build(int rt, int l, int r) {
     tree[rt].l = l, tree[rt].r = r;
     if (l == r) {
     	tree[rt].minn = w[val[l]];
     	return ;
	 }   
	 int mid = (l + r) >> 1;
	 build(lson, l, mid), build(rson, mid + 1, r);
	 push_up(rt);
   }
   void update(int rt, int l, int r, int p, int k) {
   	  if (tree[rt].l == tree[rt].r) {
   	      tree[rt].minn = k;
   	      return ;
	   }
	  int mid = (l + r) >> 1;
	  if (p <= mid) update(lson, l, mid, p, k);
	  else update(rson, mid + 1, r, p, k);
	  push_up(rt);
   }
   int query(int rt, int l, int r, int L, int R) {
   	  if (L <= l && r <= R) return tree[rt].minn;
   	  int ret = INF;
   	  int mid = (l + r) >> 1;
   	  if (L <= mid) ret = min(ret, query(lson, l, mid, L, R));
   	  if (R > mid) ret = min(ret, query(rson, mid + 1, r, L, R));
   	  return ret;
   }
}
using namespace Seg;

namespace Cut{ 
   void dfs(int x, int f) {
   	fa[x] = f, dep[x] = dep[f] + 1, siz[x] = 1; 
	for (int i = head[x]; i; i = e[i].nxt) {
   	        int v = e[i].v;
			if (v == f) continue;
			dfs(v, x);
		  siz[x] += siz[v];
		  if (siz[v] > siz[son[x]]) son[x] = v;	
	   }
   }
   void dfs2(int x, int tp) {
   	 top[x] = tp, id[x] = ++o, val[o] = x;
	 if (son[x]) dfs2(son[x], tp);
   	 for (int i = head[x]; i; i = e[i].nxt) {
   	 	    int v = e[i].v;
   	 	    if(v == fa[x] || v == son[x]) continue;
   	 	    dfs2(v, v);
		}
   } 
   int Query(int x, int y) {//查询路径最小值 
   	  int ret = INF;
   	  while(top[x] != top[y]) {
   	     if (dep[top[x]] < dep[top[y]]) swap(x, y);
		 ret = min(ret, query(1, 1, fcnt, id[top[x]], id[x]));
		 x = fa[top[x]]; 	
	   }
	  if (dep[x] > dep[y]) swap(x, y);
	  ret = min(ret, query(1, 1, fcnt, id[x], id[y]));
	  if (x > n) ret = min(ret, w[fa[x]]);
	  return ret;
   }
}


using namespace Cut;

int dfn[MAXN], low[MAXN], Stack[MAXN], tot, cnt;
void Tarjan(int x) {
   dfn[x] = low[x] = ++tot;
   Stack[++cnt] = x;
   for (int i = Head[x]; i; i = E[i].nxt) {
   	     int v = E[i].v;
   	     if (!dfn[v]) {
   	        Tarjan(v);
			low[x] = min(low[x], low[v]);
		    if (low[v] == dfn[x]) {
		       fcnt++;//方点的个数
			   for (int j = 0; j != v; cnt--) {
			   	   j = Stack[cnt];
			   	   add_edge(fcnt, j), add_edge(j, fcnt);//方点和原点建边 
			   } 
			   add_edge(fcnt, x), add_edge(x, fcnt);//把最后一个点加上 
			}
		 }
		else low[x] = min(low[x], dfn[v]);
   }
}
char s[5];
int main(){
   n = read(), m = read(), q = read();
   for (int i = 1; i <= n; i++) w[i] = read();
   for (int i = 1, u, v; i <= m; i++) {
   	   u = read(), v = read();
   	   Add_edge(u, v), Add_edge(v, u); 
   }
   fcnt = n;
   Tarjan(1), dfs(1, 0), dfs2(1, 1);
   for (int i = 1; i <= n; i++)  
      if (fa[i]) S[fa[i]].insert(w[i]);//存每个点的儿子 
   for (int i = n + 1; i <= fcnt; i++)  w[i] = *S[i].begin();//每个方点存周围点的最小值 
   build(1, 1, fcnt);
   while(q--) {
   	  scanf("%s", s + 1);
   	  if (s[1] == 'C') {
   	    int x = read(), y = read();
   	    update(1, 1, fcnt, id[x], y);
   	    if (fa[x]) {
   	       S[fa[x]].erase(S[fa[x]].lower_bound(w[x]));
		   S[fa[x]].insert(y);
		   if (w[fa[x]] != *S[fa[x]].begin()) {
		   	    w[fa[x]] = *S[fa[x]].begin();
		   	    update(1, 1, fcnt, id[fa[x]], w[fa[x]]);
		   }	
		}
		w[x] = y;
	  }
	  else if(s[1] == 'A'){
	  	 int x = read(), y = read();
	  	 printf("%d\n", Query(x, y));
	  }
   }
   puts(""); 
   return 0;
}