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 \(\text{multiset}\) 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\n", Gao::query(x, y));
}
return 0;
}