BZOJ1969 [AHIO2005]航線規劃
阿新 • • 發佈:2018-11-16
Description
給定一個無向圖,需要支援兩個操作,斷掉一條邊或者詢問兩個點之間的路徑上有多少個橋
Solution
考慮把操作離線,然後時光倒流一下。初始的時候所有的邊權值都為\(1\),把加入邊後形成的環直接置成\(0\)。詢問就直接查詢兩點之間的鏈上權值和
Code
#include <bits/stdc++.h> using namespace std; #define fst first #define snd second #define squ(x) ((LL)(x) * (x)) #define debug(...) fprintf(stderr, __VA_ARGS__) typedef long long LL; typedef pair<int, int> pii; inline int read() { int sum = 0, fg = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') fg = -1; for (; isdigit(c); c = getchar()) sum = (sum << 3) + (sum << 1) + (c ^ 0x30); return fg * sum; } const int maxn = 3e4 + 10; const int maxm = 2e5 + 10; int n, m, cnt; vector<int> g[maxn]; multiset<pii> Set; struct node { int x, y, ty; }tt[maxm]; int Fa[maxn], S[maxn]; int find(int x) { return x == Fa[x] ? x : Fa[x] = find(Fa[x]); } namespace ST { #define ls (rt << 1) #define rs (rt << 1 | 1) int A[maxn << 2], tag[maxn << 2]; void push_up(int rt) { A[rt] = A[ls] + A[rs]; } void push_down(int rt, int l, int r) { if (~tag[rt]) { tag[ls] = tag[rs] = tag[rt]; int mid = (l + r) >> 1; A[ls] = tag[rt] * (mid - l + 1); A[rs] = tag[rt] * (r - mid); tag[rt] = -1; } } void change(int rt, int l, int r, int L, int R, int v) { if (L <= l && r <= R) { A[rt] = v * (r - l + 1), tag[rt] = v; return; } push_down(rt, l, r); int mid = (l + r) >> 1; if (L <= mid) change(ls, l, mid, L, R, v); if (R > mid) change(rs, mid + 1, r, L, R, v); push_up(rt); } int query(int rt, int l, int r, int L, int R) { if (L <= l && r <= R) return A[rt]; push_down(rt, l, r); int mid = (l + r) >> 1, res = 0; if (L <= mid) res += query(ls, l, mid, L, R); if (R > mid) res += query(rs, mid + 1, r, L, R); return res; } void init() { memset(tag, -1, sizeof tag); change(1, 1, n, 1, n, 1); } } int d[maxn], fa[maxn], sz[maxn], top[maxn], hs[maxn]; int Index, dfn[maxn]; void dfs1(int now, int f) { d[now] = d[f] + 1, fa[now] = f, sz[now] = 1; int ms = 0; for (int i = 0; i < g[now].size(); i++) { int son = g[now][i]; if (son == f) continue; dfs1(son, now); sz[now] += sz[son]; if (sz[son] > sz[ms]) ms = son; } hs[now] = ms; } void dfs2(int now, int topf) { dfn[now] = ++Index; top[now] = topf; if (!hs[now]) return; dfs2(hs[now], topf); for (int i = 0; i < g[now].size(); i++) { int son = g[now][i]; if (son == fa[now] || son == hs[now]) continue; dfs2(son, son); } } void change(int x, int y, int v) { while (top[x] != top[y]) { if (d[top[x]] < d[top[y]]) swap(x, y); ST::change(1, 1, n, dfn[top[x]], dfn[x], v); x = fa[top[x]]; } if (d[x] > d[y]) swap(x, y); if (dfn[x] < dfn[y]) ST::change(1, 1, n, dfn[x] + 1, dfn[y], v); } int query(int x, int y) { int res = 0; while (top[x] != top[y]) { if (d[top[x]] < d[top[y]]) swap(x, y); res += ST::query(1, 1, n, dfn[top[x]], dfn[x]); x = fa[top[x]]; } if (d[x] > d[y]) swap(x, y); if (dfn[x] < dfn[y]) res += ST::query(1, 1, n, dfn[x] + 1, dfn[y]); return res; } int main() { freopen("line.in", "r", stdin); freopen("line.out", "w", stdout); n = read(), m = read(); for (int i = 1; i <= m; i++) { int x = read(), y = read(); if (x > y) swap(x, y); Set.insert((pii){x, y}); } int op; while (~(op = read())) { int x = read(), y = read(); if (x > y) swap(x, y); if (!op) Set.erase(Set.lower_bound((pii){x, y})); tt[++cnt] = (node){x, y, op}; } for (int i = 1; i <= n; i++) Fa[i] = i; for (multiset<pii>::iterator it = Set.begin(); it != Set.end(); ++it) { int x = it->fst, y = it->snd; if (find(x) == find(y)) { tt[++cnt] = (node){x, y, 0}; continue; } Fa[find(x)] = find(y); g[x].push_back(y), g[y].push_back(x); } dfs1(1, 0); dfs2(1, 1); ST::init(); for (int i = cnt; i >= 1; i--) { if (tt[i].ty) S[++S[0]] = query(tt[i].x, tt[i].y); else change(tt[i].x, tt[i].y, 0); } for (int i = S[0]; i >= 1; i--) printf("%d\n", S[i]); return 0; }