BZOJ1969 [AHIO2005]航线规划

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
", S[i]);

	return 0;
}