无向图的双联通分量

Posted 2020-10-12 yohanlong

点双和边双的区别我在上一篇文章中已经讨论过了，这篇文章讲边双的求法。

由于是边双，就决定了边双中一定不含有桥，但是可以含有割顶。

所以我们对边双唯一的限制条件就是不经过桥。

如此一来，我们可以分成两次dfs，第一次求出所有的桥，第二次dfs时遍历整张图，只要保证不经过桥就可以了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>

using namespace std;

const int maxn = 105, maxm = maxn * 2;

int n, m, tot, dfs_clock, bcc_cnt;

int h[maxn], dfn[maxn], low[maxn], iscut[maxn], vis[maxn];
 
struct edge
{
    int v, next, isbridge;
}a[maxm];

vector<int> bcc[maxn];

void add(int x, int y)
{
    a[tot].v = y;
    a[tot].next = h[x];
    h[x] = tot++;
}

int dfs(int u, int fa)
{
    int lowu = dfn[u] = ++dfs_clock;
    int child = 0;
    for (int i = h[u]; ~i; i = a[i].next)
    {
        int v = a[i].v;
        if (!dfn[v])
        {
            child++;
            int lowv = dfs(v, u);
            lowu = min(lowu, lowv);
            if (lowv >= dfn[u])
            {
                iscut[u] = 1;
            }
            if (lowv > dfn[u])
            {
//                printf("%d - %d\n", u, v);
                a[i].isbridge = 1;
                a[i^1].isbridge = 1;
            }
        }else if (dfn[v] < dfn[u] && v != fa)
        {
            lowu = min(lowu, dfn[v]);
        }
    }
    if (fa == 0 && child == 1)
    {
        iscut[u] = 0;
    }
    low[u] = lowu;
    return lowu;
}

void dfs2(int u, int fa)
{
    bcc[bcc_cnt].push_back(u);
    vis[u] = 1;
    for (int i = h[u]; ~i; i = a[i].next)
    {
        int v = a[i].v;
        if (v == fa || a[i].isbridge ||vis[v]) continue;
        dfs2(v, u);
    } 
}

int main()
{
    freopen("无向图双联通分量.in","r",stdin);
    scanf("%d%d", &n, &m);
    memset(h, -1, sizeof h); tot = dfs_clock = 0;
    for (int i = 1; i <= m; i++)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y); add(y, x);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; i++)
        if (!vis[i])
        {
            bcc_cnt++;
            bcc[bcc_cnt].clear();
            dfs2(i, 0);
        }
    for (int i = 1; i <= bcc_cnt; i++)
    {
        for (int j = 0; j < bcc[i].size(); j++)
            printf("%d ", bcc[i][j]);
        printf("\n");
    }
    return 0;
}