CF 962F Simple Cycles Edges

题目描述

https://codeforces.com/problemset/problem/962/F

简要题意:给定一个 $n$ 个点 $m$ 条边的简单无向图,求恰好被包含在一个简单环中的边

$n,m\le 10^5$

Solution

注意到一个简单环就是一个双连通分量,这时候我们要考虑使用边双还是点双

如果两个简单环只有一个交点,那么这两个简单环是独立的

所以我们使用点双,因为边双内可能有多个独立的简单环

对于一个点双,如果点数等于边数,那么这个点双就是一个简单环

所以我们只需要统计点数等于边数的点双即可

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#include <iostream>
#include <cstdio>
#include <stack>
#include <set>
#include <algorithm>
#include <vector>
#include <numeric>
#define maxn 100010
#define maxm 100010
using namespace std;

int n, m;

struct Edge {
int fr, to, next, ext;
} e[maxm * 2]; int c1, head[maxn];
inline void add_edge(int u, int v) {
e[c1].fr = u; e[c1].to = v;
e[c1].next = head[u]; head[u] = c1++;
}

int dfn[maxn], low[maxn], ans[maxm]; stack<int> S, E;
void tarjan(int u, int fa) {
static int cnt = 0;
dfn[u] = low[u] = ++cnt; S.push(u);
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, ext = e[i].ext; if (v == fa) continue;
int cur = E.size();
if (!ext) E.push(i), ext = e[i ^ 1].ext = 1;
if (!dfn[v]) {
tarjan(v, u); low[u] = min(low[u], low[v]);
if (low[v] >= dfn[u]) {
int t, sum = 1, ok;
do {
t = S.top(); S.pop();
++sum;
} while (t != v);
ok = sum == E.size() - cur;
while (E.size() > cur) {
ans[E.top() / 2 + 1] = ok;
E.pop();
}
}
}
else low[u] = min(low[u], dfn[v]);
}
}

int main() { fill(head, head + maxn, -1);
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);

cin >> n >> m;
for (int i = 1; i <= m; ++i) {
int x, y; cin >> x >> y;
add_edge(x, y); add_edge(y, x);
}
for (int i = 1; i <= n; ++i) if (!dfn[i]) tarjan(i, 0);
cout << accumulate(ans + 1, ans + m + 1, 0) << "\n";
for (int i = 1; i <= m; ++i)
if (ans[i]) cout << i << " ";
return 0;
}