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HDU 2460 Network

题目链接

题意：给定一个无向图，问每次加入一条边，问个图中还剩多少桥

思路：先双连通缩点，然后形成一棵树，每次加入一条边，相当于询问这两点路径上有多少条边，这个用树链剖分+线段树处理

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

#pragma comment(linker, "/STACK:1024000000,1024000000");
const int N = 100005;
const int M = 200005;

int n, m;

struct Edge {
	int u, v, id;
	bool iscut;
	Edge() {}
	Edge(int u, int v, int id) {
		this->u = u; this->v = v; 
		this->id = id; this->iscut = false;
	}
} edge[M * 2], cut[M];

int en, cn, first[N], next[M * 2];

void init() {
	en = 0;
	memset(first, -1, sizeof(first));
}

void add_edge(int u, int v, int id) {
	edge[en] = Edge(u, v, id);
	next[en] = first[u];
	first[u] = en++;
}

int pre[N], dfn[N], dfs_clock;

void dfs_cut(int u, int fa) {
	pre[u] = dfn[u] = ++dfs_clock;
	for (int i = first[u]; i + 1; i = next[i]) {
		if (edge[i].id == fa) continue;
		int v = edge[i].v;
		if (!pre[v]) {
			dfs_cut(v, edge[i].id);
			dfn[u] = min(dfn[u], dfn[v]);
			if (dfn[v] > pre[u]) {
				cut[cn++] = edge[i];
				edge[i].iscut = edge[i^1].iscut = true;
			}
		} else dfn[u] = min(dfn[u], pre[v]);
	}
}

void find_cut() {
	cn = dfs_clock = 0;
	memset(pre, 0, sizeof(pre));
	for (int i = 1; i <= n; i++)
		if (!pre[i]) dfs_cut(i, -1);
}

vector<int> g[N];
int bccno[N], bccn;

void dfs_bcc(int u) {
	bccno[u] = bccn;
	for (int i = first[u]; i + 1; i = next[i]) {
		if (edge[i].iscut) continue;
		int v = edge[i].v;
		if (bccno[v]) continue;
		dfs_bcc(v);
	}
}

void find_bcc() {
	bccn = 0;
	memset(bccno, 0, sizeof(bccno));
	for (int i = 1; i <= n; i++)
		if (!bccno[i]) {
			bccn++;
			dfs_bcc(i);
		}
	for (int i = 1; i <= bccn; i++) g[i].clear();
	for (int i = 0; i < cn; i++) {
		int u = bccno[cut[i].u], v = bccno[cut[i].v];
		g[u].push_back(v);
		g[v].push_back(u);
	}
}

int dep[N], fa[N], son[N], sz[N], top[N], id[N], idx;

void dfs1(int u, int f, int d) {
	dep[u] = d; sz[u] = 1; fa[u] = f; son[u] = 0;
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if (v == f) continue;
		dfs1(v, u, d + 1);
		sz[u] += sz[v];
		if (sz[son[u]] < sz[v])
			son[u] = v;
	}
}

void dfs2(int u, int tp) {
	id[u] = ++idx; top[u] = tp;
	if (son[u]) dfs2(son[u], tp);
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if (v == fa[u] || v == son[u]) continue;
		dfs2(v, v);
	}
}

#define lson(x) ((x<<1)+1)
#define rson(x) ((x<<1)+2)

struct Node {
	int l, r, sum, flag;
	void gao() {
		sum = 0;
		flag = 1;
	}
} node[N * 4];

void build(int l, int r, int x = 0) {
	node[x].l = l; node[x].r = r; node[x].flag = 0;
	node[x].sum = r - l + 1;
	if (l == r) return;
	int mid = (l + r) / 2;
	build(l, mid, lson(x));
	build(mid + 1, r, rson(x));
}

void pushup(int x) {
	node[x].sum = node[lson(x)].sum + node[rson(x)].sum;
}

void pushdown(int x) {
	if (node[x].flag) {
		node[lson(x)].gao();
		node[rson(x)].gao();
		node[x].flag = 0;
	}
}

int query(int l, int r, int x = 0) {
	if (node[x].l >= l && node[x].r <= r) {
		int ans = node[x].sum;
		node[x].gao();
		return ans;
	}
	pushdown(x);
	int mid = (node[x].l + node[x].r) / 2;
	int ans = 0;
	if (l <= mid) ans += query(l, r, lson(x));
	if (r > mid) ans += query(l, r, rson(x));
	pushup(x);
	return ans;
}

int gao(int u, int v) {
	int tp1 = top[u], tp2 = top[v];
	int ans = 0;
	while (tp1 != tp2) {
		if (dep[tp1] < dep[tp2]) {
			swap(tp1, tp2);
			swap(u, v);
		}
		ans += query(id[tp1], id[u]);
		u = fa[tp1];
		tp1 = top[u];
	}
	if (u == v) return ans;
	if (dep[u] > dep[v]) swap(u, v);
	ans += query(id[son[u]], id[v]);
	return ans;
}

int main() {
	int cas = 0;
	while (~scanf("%d%d", &n, &m) && n || m) {
		init();
		int u, v;
		for (int i = 0; i < m; i++) {
			scanf("%d%d", &u, &v);
			add_edge(u, v, i);
			add_edge(v, u, i);
		}
		find_cut();
		find_bcc();
		idx = 0;
		dfs1(1, 0, 1);
		dfs2(1, 1);
		int q;
		scanf("%d", &q);
		printf("Case %d:\n", ++cas);
		build(1, n);
		while (q--) {
			scanf("%d%d", &u, &v);
			u = bccno[u]; v = bccno[v];
			cn -= gao(u, v);
			printf("%d\n", cn);
		}
		printf("\n");
	}
	return 0;
}

HDU 2460 Network（双连通+树链剖分+线段树）

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原文地址：http://blog.csdn.net/accelerator_/article/details/40395847