UOJ #30. [CF Round #278] Tourists

UOJ #30. [CF Round #278] Tourists

题目大意 :
有一张 (n) 个点， (m) 条边的无向图，每一个点有一个点权 (a_i) ，你需要支持两种操作，第一种操作修改一个点的点权，第二种操作给出 (u, v)，求一个点 (x) ，存在一条 (u-x-v) 不经过重复点的路径且 (a_x) 最小

(1 leq n,m leq 10^5)

解题思路 :

​ 考虑如果 (u-x) 和 (x-v) 经过了同一个割点，那么一定不合法，也就是说 (x) 一定在 (u-v) 之间的点双联通分量里。于是将圆方树建出来，对于每一个方点用一个 ( ext{multiset}) 维护最小值，树链剖分维护路径答案，注意特判 (lca) 是方点的情况

/*program by mangoyang*/
#include<bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
	int f = 0, ch = 0; x = 0;
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
	for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
	if(f) x = -x;
}

#define lson (u << 1)
#define rson (u << 1 | 1)

const int N = 1000005;
char str[20];
vector<int> g[N];
int ass[N], n, m, q;

struct SegmentTree{
	int s[N<<2];
	inline SegmentTree(){ memset(s, 127, sizeof(s)); }
	inline void modify(int u, int l, int r, int pos, int x){
		if(l == r) return (void) (s[u] = x);
		int mid = l + r >> 1;
		if(pos <= mid) modify(lson, l, mid, pos, x);
		else modify(rson, mid + 1, r, pos, x);
		s[u] = min(s[lson], s[rson]);
	}
	inline int query(int u, int l, int r, int L, int R){
		if(l >= L && r <= R) return s[u];
		int mid = l + r >> 1, res = inf;
		if(L <= mid) res = min(query(lson, l, mid, L, R), res);
		if(mid < R) res = min(query(rson, mid + 1, r, L, R), res);
		return res;
	}
}Seg;
namespace Gao{
	multiset<int> st[N];
	vector<int> g[N];
	int dep[N], sz[N], ff[N], ms[N], dfn[N], top[N], cnt;
	inline void dfs(int u, int fa){	
		dep[u] = dep[fa] + 1, sz[u] = 1, ff[u] = fa;
		if(fa > n) st[fa].insert(ass[u]);
		for(int i = 0; i < g[u].size(); i++){
			int v = g[u][i];
			if(v == fa) continue;
			dfs(v, u), sz[u] += sz[v];
			if(sz[v] > sz[ms[u]]) ms[u] = v;
		}
	}
	inline void split(int u, int chain, int fa){
		dfn[u] = ++cnt, top[u] = chain;
		int tmp = u <= n ? ass[u] : (*st[u].begin());
		Seg.modify(1, 1, 2 * n, dfn[u], tmp);
		if(ms[u]) split(ms[u], chain, u);
		for(int i = 0; i < g[u].size(); i++){
			int v = g[u][i];
			if(v == fa || v == ms[u]) continue;
			split(v, v, u);
		}
	}
	inline int query(int x, int y){
		int res = inf;
		while(top[x] != top[y]){
			if(dep[top[x]] < dep[top[y]]) swap(x, y);
			res = min(res, Seg.query(1, 1, 2 * n, dfn[top[x]], dfn[x]));
			x = ff[top[x]];
		}
		if(dfn[x] > dfn[y]) swap(x, y);
		res = min(res, Seg.query(1, 1, 2 * n, dfn[x], dfn[y]));
		if(ff[x] <= n && ff[x]) res = Min(res, ass[ff[x]]);
		return res;
	}
	inline void modify(int x, int y){
		Seg.modify(1, 1, 2 * n, dfn[x], y);
		if(x > 1){
			st[ff[x]].erase(st[ff[x]].find(ass[x]));
			st[ff[x]].insert(y);
			Seg.modify(1, 1, 2 * n, dfn[ff[x]], *(st[ff[x]].begin()));
		}	
		ass[x] = y;
	}
}
namespace Graph{
	int a[N], nxt[N], head[N], dfn[N], low[N], st[N], top, Index, id, cnt = 1;
	inline void add(int x, int y){
		a[++cnt] = y, nxt[cnt] = head[x], head[x] = cnt;
	}
	inline void tarjan(int u, int fa){
		dfn[u] = low[u] = ++Index;		
		for(int p = head[u]; p; p = nxt[p]){
			if((p ^ 1) == fa) continue;
			int v = a[p];
			if(dfn[v]){ low[u] = min(low[u], dfn[v]); continue; }
			st[++top] = v, tarjan(v, p);
			low[u] = min(low[u], low[v]);
			if(low[v] >= dfn[u]){
				id++;
				for(; st[top] != v; top--){
					Gao::g[id].push_back(st[top]);
					Gao::g[st[top]].push_back(id);
				}
				top--;
				Gao::g[id].push_back(v), Gao::g[v].push_back(id);
				Gao::g[id].push_back(u), Gao::g[u].push_back(id);
			}
		}
	}
	inline void Build(){
		id = n;
		for(int i = 1; i <= n; i++) if(!dfn[i]) tarjan(i, 0);
	}
}
signed main(){
	read(n), read(m), read(q);
	for(int i = 1; i <= n; i++) read(ass[i]);
	for(int i = 1, x, y; i <= m; i++){
		read(x), read(y);	
		Graph::add(x, y), Graph::add(y, x);
	}
	Graph::Build();
	Gao::dfs(1, 0), Gao::split(1, 1, 0);
	for(int i = 1, x, y; i <= q; i++){
		scanf("%s", str); read(x), read(y);
		if(str[0] == 'C') Gao::modify(x, y);
		if(str[0] == 'A') printf("%d
",  Gao::query(x, y));
	}
	return 0;
}