tarjan求边双连通分量模板

Posted 2021-08-25 hesorchen

边双连通图

不存在割边（桥）的无向连通图称为边双连通图

求解

先对原图求出割边，然后对删除割边后的图跑DFS染色即可。

代码

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
using namespace std;
int n, m;

const int N = 1010;
vector<int> edge[N];
map<int, map<int, int>> mp;
int fa[N], dfn[N], low[N], ct;
void tarjan(int u)
{
    dfn[u] = low[u] = ++ct;
    for (int i = 0; i < edge[u].size(); i++)
    {
        int v = edge[u][i];
        if (!dfn[v])
        {
            fa[v] = u;
            tarjan(v);
            low[u] = min(low[u], low[v]);
            if (low[v] > low[u])
                mp[u][v] = mp[v][u] = 1; //u-v 是桥
        }
        else if (v != fa[u])
        {
            low[u] = min(low[u], dfn[v]);
        }
    }
}
int cnt;
int belong[N];
void dfs(int u)
{
    belong[u] = cnt;//染色
    for (int i = 0; i < edge[u].size(); i++)
    {
        int v = edge[u][i];
        if (mp[u][v])
            continue;
        if (dfn[v])
        {
            dfn[v] = 0;
            dfs(v);
        }
    }
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= m; i++)
    {
        int u, v;
        cin >> u >> v;
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    tarjan(1); //求桥
    for (int i = 1; i <= n; i++)
    {
        if (dfn[i])
        {
            ++cnt;
            dfn[i] = 0;
            dfs(i); //边双缩点
        }
    }
    for (int i = 1; i <= n; i++)
        cout << i << " belong " << belong[i] << endl;
    return 0;
}