Tarjan求强连通分量求桥和割点模板

Posted 2021-01-04 rubbishes

Tarjan 求强连通分量模板、参考博客

#include<stdio.h>
#include<stack>
#include<algorithm>
using namespace std;
const int maxn = 1e3 + 10;
const int maxm = 330000 + 10;
struct EDGE{ int v, nxt; }Edge[maxm];
int Head[maxn], cnt;
int DFN[maxn], LOW[maxn], color[maxn], INDEX, id;
bool vis[maxn];
int N, M;
stack<int> stk;

inline void init()
{
    while(!stk.empty()) stk.pop();
    for(int i=0; i<=N; i++)
        Head[i] = DFN[i] = LOW[i] = color[i] = -1,
    cnt = INDEX = id = 0;
}

inline void AddEdge(int from, int to)
{
    Edge[cnt].v = to;
    Edge[cnt].nxt = Head[from];
    Head[from] = cnt++;
}

inline void tarjan(int u)
{
    DFN[u] = LOW[u] = INDEX++;
    stk.push(u);
    vis[u] = true;
    for(int i=Head[u]; i!=-1; i=Edge[i].nxt){
        int Eiv = Edge[i].v;
        if(DFN[Eiv] == -1){
            tarjan(Eiv);
            LOW[u] = min(LOW[u], LOW[Eiv]);
        }else{
            if(vis[Eiv])
                LOW[u] = min(LOW[u], LOW[Eiv]);
        }
    }

    if(DFN[u] == LOW[u]){
        color[u] = ++id;
        vis[u] = false;
        while(stk.top() != u){
            vis[stk.top()] = false;
            color[stk.top()] = id;
            stk.pop();
        }
        stk.pop();
    }
}

int main(void)
{
    while(~scanf("%d %d", &N, &M)){
        init();
        int from, to;
        while(M--){
            scanf("%d %d", &from, &to);
            AddEdge(from, to);
        }

        for(int i=0; i<N; i++)
            if(DFN[i] == -1)
                tarjan(i);

        printf("%d

", id);
    }
    return 0;
}

View Code

 

Tarjan 求桥和割点模板

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 10;///图中顶点的数量
const int maxm = 4e5 + 10;///图中边的数量
struct EDGE{ int v, nxt; }Edge[maxm];
int Head[maxn], cnt;///表头以及边的编号
int LOW[maxn], DFN[maxn];///每个点最早可回溯到的祖先节点、每个点的遍历序号
int Fa[maxn], INDEX;///Fa数组记录每一个点的父亲、INDEX是算法里的时间戳
int N, M;

inline void init()
{
    for(int i=0; i<=N; i++)
        Head[i] = LOW[i] = DFN[i] = -1, Fa[i] = 0;
    cnt = INDEX = 0;
}

inline void AddEdge(int from, int to)
{
    Edge[cnt].v = to;
    Edge[cnt].nxt = Head[from];
    Head[from] = cnt++;
}

void Tarjan(int v, int Father)
{
    Fa[v] = Father;
    DFN[v] = LOW[v] = INDEX++;
    for(int i=Head[v]; i!=-1; i=Edge[i].nxt){
        int Eiv = Edge[i].v;
        if(DFN[Eiv] == -1){
            Tarjan(Eiv, v);
            LOW[v] = min(LOW[v], LOW[Eiv]);
        }
        else if(Father != Eiv)
            LOW[v] = min(LOW[v], DFN[Eiv]);
    }
}

///这份代码中顶点编号是从 0 ~ N-1
void Count()///统计割点和桥
{
    Tarjan(0, -1);

    int Cut_Num = 0;///割点的数量
    int Root_Child  = 0;///根节点的儿子
    Tarjan(0, -1);
    for(int i=1; i<N; i++){
        int v = Fa[i];
        if(v == 0) Root_Child++;
        else if(LOW[i] >= DFN[v] && !is_cut[v])
            is_cut[v] = true, Cut_Num++;
    }
    if(Root_Child > 1)
        is_cut[0] = true, Cut_Num++;///根节点有超过一个儿子则说明也是割点

    for(int i=0; i<N; i++){
        int v = Fa[i];
        if(v >= 0 && LOW[i] > DFN[v]){
            // v->i is bridge
            //可以用一个 pair<int, int> 来记录
        }
    }
}

int main(void)
{
    while(~scanf("%d %d", &N, &M)){
        init();
        int from, to;
        while(M--){
            scanf("%d %d", &from, &to);
            AddEdge(from, to);
            AddEdge(to, from);
        }
        Count();
    }
    return 0;
}

View Code