HDU 5458 Stability（双连通分量+LCA+并查集+树状数组）（2015 ACM/ICPC Asia Regional Shenyang Online）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5458

[align=left]Problem Description[/align]
Given an undirected connected graph G with n nodes and m edges, with possibly repeated edges and/or loops. The stability of connectedness between node u and node v is defined by the number of edges in this graph which determines the connectedness between them (once we delete this edge, node u and v would be disconnected).

You need to maintain the graph G, support the deletions of edges (though we guarantee the graph would always be connected), and answer the query of stability for two given nodes.

[align=left]Input[/align]
There are multiple test cases(no more than 3 cases), and the first line contains an integer t, meaning the totally number of test cases.

For each test case, the first line contains three integers n, m and q, where 1≤n≤3×104,1≤m≤105 and 1≤q≤105. The nodes in graph G are labelled from 1 to n.

Each of the following m lines contains two integers u and v describing an undirected edge between node u and node v.

Following q lines - each line describes an operation or a query in the formats:
⋅ 1 a b: delete one edge between a and b. We guarantee the existence of such edge.
⋅ 2 a b: query the stability between a and b.

[align=left]Output[/align]
For each test case, you should print first the identifier of the test case.

Then for each query, print one line containing the stability between corresponding pair of nodes.

题目大意：给一个N个点M条边的无向图，有Q个询问：1、删掉a、b之间所存在的边；2、询问有多少条边，单独删掉之后a与b不再连通。

思路：脑洞大开。

对于询问，首先想到的就是a与b之间有多少桥（割边），然后想到双连通分量，然而删边是个坑爹的问题，于是我们离线倒着来，把删边变成加边。

双连通分量这种东西呢，其实缩点连起来之后，就是一棵树辣。

然后询问两个点的时候，设根到点x的距离为dep[x]，a、b的最近公共祖先为lca(a, b)，那么询问query(a, b) = dep[a] + dep[b] - 2 * dep[lca(a, b)]

加一条边的时候呢，比如加edge(a, b)，那么原来的a到b的路径就形成了一个环，那么这个环就应该缩成一个双连通分量（每个点只会被缩一次，平摊的复杂度肯定是没问题的）。

这里可以用并查集来维护同一个双连通分量，每次都是儿子合并到父亲，然后每一次点u合并到父亲的时候，u和u的所有子孙的高度都会减一。

因为要处理一棵子树的所有值，这里使用DFS序+树状数组的方法来维护每个点的深度dep。

然后求LCA这里用的是树上倍增。整个题目要做的就是这些了。

然后整个流程就是：

1、初始化边表并读入所有数据并给边表排序用于查找（我这里用了vector）。（复杂度O(n+m+q+mlog(m))）

2、然后给1操作涉及的边打个删除标记。（复杂度O(qlog(m))）

3、DFS随便建一棵树，给DFS到的边打个删除标记（我直接把父边从vector移除了），顺便建立好DFS序和树状数组、dep和fa数组（用于LCA倍增）。（复杂度O(n+m+nlog(n))）

4、初始化LCA倍增。（复杂度O(nlog(n))）

5、对于每条没打标记的边（即树里的横向边？） edge(a, b)，合并路径path(a, b)上的所有点，并维护好树状数组。（复杂度为O(m+nlog(n))）

6、逆序跑询问，第一个操作就加边，加边方法同流程4，第二个操作便是求出LCA，然后用树状数组扒出每个点的深度，然后加加减减得到结果并存起来。（复杂度O(qlog(n)+nlog(n)))）

7、输出结果。（复杂度O(q)）

代码（889MS）：

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <cctype>
using namespace std;
typedef long long LL;

const int MAXV = 30010;
const int MAXQ = 100010;
const int MAX_LOG = 16;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;

int readint() {
char c = getchar();
while(!isdigit(c)) c = getchar();
int res = 0;
while(isdigit(c)) res = res * 10 + c - '0', c = getchar();
return res;
}

struct Node {
int to, del;
Node(int to): to(to), del(0) {}
bool operator < (const Node &rhs) const {
if(to != rhs.to) return to < rhs.to;
return del > rhs.del;
}
};
vector<Node> adjs[MAXV];
int n, m, q, T;

void init() {
for(int i = 1; i <= n; ++i) {
adjs[i].clear();
}
}

void add_edge(int u, int v) {
adjs[u].push_back(Node(v));
adjs[v].push_back(Node(u));
}

struct Query {
int op, a, b, apos, bpos;
void read() {
scanf("%d%d%d", &op, &a, &b);
if(op == 1) {
lower_bound(adjs[a].begin(), adjs[a].end(), Node(b))->del = true;
lower_bound(adjs[b].begin(), adjs[b].end(), Node(a))->del = true;
}
}
} query[MAXQ];
int ans[MAXQ], acnt;

struct BIT {
int tree[MAXV];
void init() {
memset(tree + 1, 0, n * sizeof(int));
}
int lowbit(int x) {
return x & -x;
}
void modify(int x, int val) {
while(x <= n) {
tree[x] += val;
x += lowbit(x);
}
}
void modify(int a, int b, int val) {
modify(a, val);
modify(b + 1, -val);
}
int get_val(int x) {
int res = 0;
while(x) {
res += tree[x];
x -= lowbit(x);
}
return res;
}
} bitree;

int bid[MAXV], eid[MAXV], dep[MAXV];
int fa[MAX_LOG][MAXV];
int dfs_clock;

void dfs_id(int u, int f, int depth) {
if(f > 0) adjs[u].erase(lower_bound(adjs[u].begin(), adjs[u].end(), Node(f)));
fa[0][u] = f;
dep[u] = depth;
bid[u] = ++dfs_clock;
for(Node& p : adjs[u]) if(!p.del && !bid[p.to]) {
p.del = 1;
dfs_id(p.to, u, depth + 1);
}
eid[u] = dfs_clock;
bitree.modify(bid[u], eid[u], 1);
}
void bit_init() {
memset(bid + 1, 0, n * sizeof(int));
bitree.init();
dfs_clock = 0;
dfs_id(1, 0, 0);
}

struct LCA {
void init_lca() {
for(int k = 0; k + 1 < MAX_LOG; ++k) {
for(int u = 1; u <= n; ++u) {
if(fa[k][u] == -1) fa[k + 1][u] = -1;
else fa[k + 1][u] = fa[k][fa[k][u]];
}
}
}

int ask(int u, int v) {
if(dep[u] < dep[v]) swap(u, v);
for(int k = 0; k < MAX_LOG; ++k) {
if((dep[u] - dep[v]) & (1 << k)) u = fa[k][u];
}
if(u == v) return u;
for(int k = MAX_LOG - 1; k >= 0; --k) {
if(fa[k][u] != fa[k][v])
u = fa[k][u], v = fa[k][v];
}
return fa[0][u];
}
} lca;

int dsu[MAXV];

int find_set(int x) {
return dsu[x] == x ? x : dsu[x] = find_set(dsu[x]);
}

void mergeFa(int u, int gold) {
u = find_set(u);
while(u != gold) {
int t = find_set(fa[0][u]);
dsu[u] = t;
bitree.modify(bid[u], eid[u], -1);
u = t;
}
}

void merge(int u, int v) {
int l = find_set(lca.ask(u, v));
mergeFa(u, l);
mergeFa(v, l);
}

void init_tree() {
for(int i = 1; i <= n; ++i)
dsu[i] = i;
for(int u = 1; u <= n; ++u)
for(Node p : adjs[u]) if(!p.del) {
merge(u, p.to);
}
}

void solve() {
bit_init();
lca.init_lca();
init_tree();
for(int i = q - 1; i >= 0; --i) {
if(query[i].op == 1) {
merge(query[i].a, query[i].b);
} else {
int l = lca.ask(query[i].a, query[i].b);
ans[acnt++] = bitree.get_val(bid[query[i].a]) + bitree.get_val(bid[query[i].b]) - 2 * bitree.get_val(bid[l]);
}
}
for(int i = acnt - 1; i >= 0; --i)
printf("%d\n", ans[i]);
}

int main() {
scanf("%d", &T);
for(int t = 1; t <= T; ++t) {
scanf("%d%d%d", &n, &m, &q);
init();
for(int i = 0, u, v; i < m; ++i) {
u = readint(), v = readint();
add_edge(u, v);
}
for(int i = 1; i <= n; ++i)
sort(adjs[i].begin(), adjs[i].end());

acnt = 0;
for(int i = 0; i < q; ++i)
query[i].read();

printf("Case #%d:\n", t);
solve();
}
}

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