72 lines
2.0 KiB
C++
72 lines
2.0 KiB
C++
#include <bits/stdc++.h>
|
||
|
||
using namespace std;
|
||
//此题是P8436的弱化版本，弱化了记录每个边双中点有哪些:vector<int> belong[N];
|
||
const int N = 500010, M = 2000010 * 2;
|
||
|
||
int n, m;
|
||
|
||
int dfn[N], low[N], ts, stk[N], top;
|
||
vector<int> belong[N]; //边双中有哪些原始点
|
||
|
||
int id[N], dcc_cnt; //原始点x属于哪个边双连通分量，dcc_cnt指边双连通分量个数
|
||
int is_bridge[M]; //记录哪些边是割边
|
||
|
||
//链式前向星
|
||
int e[M], h[N], idx, w[M], ne[M];
|
||
void add(int a, int b, int c = 0) {
|
||
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
|
||
}
|
||
|
||
//边双连通分量
|
||
void tarjan(int u, int fa) {
|
||
dfn[u] = low[u] = ++ts;
|
||
stk[++top] = u;
|
||
for (int i = h[u]; ~i; i = ne[i]) {
|
||
int j = e[i];
|
||
if (!dfn[j]) {
|
||
tarjan(j, i);
|
||
low[u] = min(low[u], low[j]);
|
||
if (dfn[u] < low[j]) is_bridge[i] = is_bridge[i ^ 1] = true; //记录割边
|
||
} else if (i != (fa ^ 1))
|
||
low[u] = min(low[u], dfn[j]);
|
||
}
|
||
|
||
if (dfn[u] == low[u]) {
|
||
++dcc_cnt; //边双数量+1
|
||
int x;
|
||
do {
|
||
x = stk[top--];
|
||
id[x] = dcc_cnt; // 记录点与边双关系
|
||
belong[dcc_cnt].push_back(x); // 记录边双中有哪些点
|
||
} while (x != u);
|
||
}
|
||
}
|
||
|
||
int main() {
|
||
scanf("%d %d", &n, &m);
|
||
memset(h, -1, sizeof h);
|
||
|
||
for (int i = 0; i < m; i++) {
|
||
int a, b;
|
||
scanf("%d %d", &a, &b);
|
||
add(a, b), add(b, a);
|
||
}
|
||
|
||
for (int i = 1; i <= n; i++)
|
||
if (!dfn[i]) tarjan(i, -1);
|
||
|
||
//输出边双个数
|
||
printf("%d\n", dcc_cnt);
|
||
|
||
// for (int i = 1; i <= dcc_cnt; i++) {
|
||
// //此边双中点的数量
|
||
// printf("%d ", belong[i].size());
|
||
// //此边双中都有哪些点
|
||
// for (int j = 0; j < belong[i].size(); j++)
|
||
// printf("%d ", belong[i][j]);
|
||
// puts("");
|
||
// }
|
||
|
||
return 0;
|
||
} |