强连通分量 & 割点/桥 & 点/边双连通分量 [模板]_点双连通分量和割点百度文库-优快云博客

本文深入探讨了图论中的关键算法，包括强连通分量、割点、点双连通分量、割边及边双连通分量的Tarjan算法实现，并提供了详细的代码示例。

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#include<stdio.h>
#include<stack>
using namespace std;
const int maxn = 1e5 + 5;
stack<int> s;
bool iscut[maxn];
int idx, cnt, btot;
int h[maxn];
int low[maxn], dfn[maxn];
int scc[maxn], bcc[maxn];
struct edge
{
    int nxt, v;
}e[maxn * 2];
//强连通分量 
void tarjan(int u)
{
    low[u] = dfn[u] = ++ idx;
    s.push(u);
    for (int i = h[u]; i; i = e[i].nxt)
    {
        int v = e[i].v;
        if(!dfn[v]) tarjan(v), low[u] = min(low[u], low[v]);
        else if(!scc[v]) low[u] = min(low[u], dfn[v]);
    }
    if(low[u] == dfn[u])
    {
        cnt ++;
        while(true)
        {
            int x = s.top(); 
            scc[x] = cnt, s.pop(); 
            if(x == u) break;
        }
    }
}

//割点
void tarjan(int u, int fa)
{
    int child = 0;
    low[u] = dfn[u] = ++ idx;
    for (int i = h[u]; i; i = e[i].nxt)
    {
        int v = e[i].v;
        if(!dfn[v])
        {
            child ++;
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
            if(low[v] >= dfn[u]) iscut[u] = true;
        }
        else if(dfn[v] < dfn[u] && v != fa)
            low[u] = min(low[u], dfn[v]);
    }
    if(child == 1 && fa < 0) iscut[u] = false;
}

//点双连通分量
void tarjan(int u, int fa)
{
    dfn[u] = low[u] = ++ idx;
    for (int i = h[u]; i; i = e[i].nxt)
    {
        int v = e[i].v;
        if(!dfn[v])
        {
            s.push(i);
            tarjan(v, u);
            if(low[v] >= low[u]) 
            {
                btot ++;
                while(true)
                {
                    int x = s.top(); s.pop();
                    bcc[x] = bcc[x ^ 1] = btot;
                    if(x == i) break;
                }
                //another edition
                 /*-----------------------------------------------------------------------
                while(1){
                    int x = sta[topp--] ;
                    if(bccno[ P[x].fr ] != Bcc_cnt){
                        bcc[ Bcc_cnt ].push_back( P[x].fr ) ; bccno[ P[x].fr ] = Bcc_cnt ;
                    }
                    if(bccno[ P[x].to ] != Bcc_cnt){
                        bcc[ Bcc_cnt ].push_back( P[x].to ) ; bccno[ P[x].to ] = Bcc_cnt ;
                    }
                    if(x == i) break ;
                }
                ------------------------------------------------------------------------*/
            }
        }
        else if(dfn[v] < dfn[u] && v != fa)
            s.push(i),
            low[u] = min(low[u], dfn[v]);
    } 
}

//割边
int num = 1;
void tarjan(int u, int fa)
{
    int child = 0;
    low[u] = dfn[u] = ++ idx;
    for (int i = h[u]; i; i = e[i].nxt)
    {
        int v = e[i].v;
        if(!dfn[v])
        {
            child ++;
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
            if(low[v] > dfn[u]) iscut[i] = iscut[i ^ 1] = true;
        }
        else if(dfn[v] < dfn[u] && v != fa)
            low[u] = min(low[u], dfn[v]);
    }
}

//边双连通分量 同强连通分量 
void tarjan(int u, int fa)
{

    for (int i = h[u]; i; i = e[i].nxt)
    {
        int v = e[i].v;
        if(!dfn[v]) tarjan(v, u), low[u] = min(low[u], low[v]);
        else if(dfn[v] < dfn[u] && v != fa)
            low[u] = min(low[u], dfn[v]);
    }
    if(low[u] == dfn[u])
    {
        cnt ++;
        while(true)
        {
            int x = s.top();
            bcc[x] = cnt, s.pop();
            if(x == u) break;
        }
    }
}
/*----------------------------------------
同样可以标记出所有桥， 然后dfs不经过桥即可
----------------------------------------*/
int main(){}