题目蓝链

Description

给定一个无向图，需要支持两个操作，断掉一条边或者询问两个点之间的路径上有多少个桥

Solution

考虑把操作离线，然后时光倒流一下。初始的时候所有的边权值都为\(1\)，把加入边后形成的环直接置成\(0\)。询问就直接查询两点之间的链上权值和

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;
}