UOJ 30 【CF Round #278】Tourists
阿新 • • 發佈:2019-01-04
Tourists
題目大意
\(n\) 個 \(m\) 條邊的無向圖,每個點有點權 \(w_i\) , \(q\) 次詢問,每次修改第 \(a\) 個點的點權為 \(w\) ,或者查詢 \(a\) 到 \(b\) 所有路徑中,最小的點權
資料範圍
\(1 \le n,m,q \le 10^5, 1 \le w_i \le 10^9\)
時空限制
2s,256MB
分析
對於這種在無向圖上查詢路徑資訊的題,一般利用圓方樹將其轉為樹上問題
對於一個方點,維護點雙中最小的點權,但是如果這樣,每修改一次就需要修改所有與它相鄰的節點。
對此我們有一個經典方法,對於方點維護除了環頂之外的資訊,在圓方樹中就相當於方點維護所有它兒子的資訊,那麼我們修改時只需要修改它的父親,查詢的時候如果 \(lca\)
Code
#include <cstdio> #include <cstring> #include <iostream> #include <vector> #include <set> using namespace std; template<class T> void read(T & x) { x = 0; int f = 1, ch = getchar(); while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9') { x = x * 10 - '0' + ch; ch = getchar(); } x *= f; } #define lson u << 1, l, mid #define rson u << 1 | 1, mid + 1, r const int inf = 1000000000; const int maxn = 100000 + 5; const int maxm = 100000 + 5; const int maxe = maxm * 2; const int maxnode = maxn * 2; int n, m, q; int val[maxnode]; struct edge { int to, nex; edge(int to = 0, int nex = 0) : to(to), nex(nex) {} } g[maxe]; int head[maxn]; int ecnt; vector<int> adj[maxnode]; inline void addedge(int u, int v) { g[ecnt] = edge(v, head[u]), head[u] = ecnt++; g[ecnt] = edge(u, head[v]), head[v] = ecnt++; } inline void adde(int u, int v) { adj[u].push_back(v); adj[v].push_back(u); } int dfc, dfn[maxn], low[maxn]; int top, sta[maxn]; int bcnt; multiset<int> bccval[maxn]; void tarjan(int u, int fa) { dfn[u] = low[u] = ++dfc; sta[++top] = u; for(int i = head[u]; ~ i; i = g[i].nex) { int v = g[i].to; if(v != fa) { if(!dfn[v]) { tarjan(v, u); low[u] = min(low[u], low[v]); if(low[v] >= dfn[u]) { int now = n + (++bcnt); adde(now, u); while(true) { int x = sta[top--]; bccval[bcnt].insert(val[x]); adde(now, x); if(x == v) break; } val[now] = * bccval[bcnt].begin(); } } else low[u] = min(low[u], dfn[v]); } } } struct segment_tree { int mn[maxnode << 2]; inline void pushup(int u) { mn[u] = min(mn[u << 1], mn[u << 1 | 1]); } void build(int u, int l, int r) { if(l == r) { mn[u] = val[ver[l]]; return; } int mid = (l + r) >> 1; build(lson), build(rson); pushup(u); } void update(int u, int l, int r, int qp) { if(l == r) { mn[u] = val[ver[l]]; return; } int mid = (l + r) >> 1; if(qp <= mid) update(lson, qp); else update(rson, qp); pushup(u); } int query(int u, int l, int r, int ql, int qr) { if(l == ql && r == qr) return mn[u]; int mid = (l + r) >> 1; if(qr <= mid) return query(lson, ql, qr); else if(ql > mid) return query(rson, ql, qr); else { int lv = query(lson, ql, mid); int rv = query(rson, mid + 1, qr); return min(lv, rv); } } int dep[maxnode], anc[maxnode], siz[maxnode], son[maxnode]; int dfn[maxnode], top[maxnode], ver[maxnode], dfc; void dfs1(int u) { siz[u] = 1; for(unsigned int i = 0; i < adj[u].size(); ++i) { int v = adj[u][i]; if(v != anc[u]) { dep[v] = dep[u] + 1; anc[v] = u; dfs1(v); siz[u] += siz[v]; if(siz[son[u]] <= siz[v]) son[u] = v; } } } void dfs2(int u, int chain) { top[u] = chain; dfn[u] = ++dfc; ver[dfc] = u; if(son[u]) { dfs2(son[u], chain); } for(unsigned int i = 0; i < adj[u].size(); ++i) { int v = adj[u][i]; if(v != anc[u] && v != son[u]) { dfs2(v, v); } } } void init() { dfs1(1); dfs2(1, 1); build(1, 1, n + bcnt); } void update(int u, int x) { int v = anc[u]; if(v) bccval[v - n].erase(bccval[v - n].find(val[u])); val[u] = x; if(v) bccval[v - n].insert(val[u]); if(v) val[v] = * bccval[v - n].begin(); update(1, 1, n + bcnt, dfn[u]); if(v) update(1, 1, n + bcnt, dfn[v]); } int query(int u, int v) { int re = inf; while(top[u] != top[v]) { if(dep[top[u]] > dep[top[v]]) swap(u, v); re = min(re, query(1, 1, n + bcnt, dfn[top[v]], dfn[v])); v = anc[top[v]]; } if(dep[u] > dep[v]) swap(u, v); re = min(re, query(1, 1, n + bcnt, dfn[u], dfn[v])); if(u > n) { re = min(re, val[anc[u]]); } return re; } } seg; void init() { tarjan(1, 0); seg.init(); } int main() { // freopen("testdata.in", "r", stdin); read(n), read(m), read(q); for(int i = 1; i <= n; ++i) { read(val[i]); } memset(head, -1, sizeof(head)); for(int i = 1; i <= m; ++i) { int u, v; read(u), read(v); addedge(u, v); } init(); for(int i = 1; i <= q; ++i) { char op[5]; scanf("%s", op); if(op[0] == 'C') { int a, w; read(a), read(w); seg.update(a, w); } else { int a, b; read(a), read(b); printf("%d\n", seg.query(a, b)); } } return 0; }