About Tarjan

I really pig brain wow

Sister school is all about is I ate

moment

Today I review a bit, think it is better to write it down

After all, my memory

Like the winter wind

Not only brush brush also blow whiz

 

About a cut point and cutting edge (Bridge):

Cutpoint : FIG puncturing it and even to become non-communication side after

Can become a point of cutting conditions : 1. For the root node, there are two or more sub-tree 2. For non-root non-leaf nodes of a subtree does not point back to the edge of the ancestors of u

Cutting edge : Deletion After this edge becomes not communicate FIG.

Cutting edge condition becomes : (u, v) and the side of the tree when low [v]> dfn [u ] Reason: v represents only nodes connected by this edge and u

Example: Luo Gu P3388 [template] point cut (top cut)

Code:

#include <iostream>
#include <cstdio>
#include <cstring>
#define N 100010
using namespace std;
struct node {
    int next, to;
}e[N * 2];
int n, m, idx, cnt, tot;
int head[N], dfn[N], low[N];
bool cut[N];
void add (int x, int y) {
    e[++cnt].next = y;
    e[cnt].to = head[x];
    head[x] = cnt;
}
void tarjan (int u, int fa) {
    dfn[u] = low[u] = ++idx;
    int child = 0;
     for (int i = head[u]; i; i = e[i].to) {
         int nx = e[i].next;
         if (!dfn[nx]) {
             tarjan (nx, fa);
             low[u] = min (low[u], low[nx]);
             if (low[nx] >= dfn[u] && u != fa) 
                 cut[u] = 1;
             if (u == fa) child++;
         }
        else low[u] = min (low[u], dfn[nx]);
    }
    if (child >= 2 && u == fa)
        cut[u] = 1;
}
int main () {
    scanf ("%d%d", &n, &m);
    for (int i = 1; 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, i);
    for (int i = 1; i <= n; i++) 
        if (cut[i])
            tot++;
    printf ("%d\n", tot);
    for (int i = 1; i <= n; i++)
        if (cut[i])
            printf ("%d ", i);
    return 0;
}

code

 

 

About strongly connected components:

Strong maximum Unicom subgraph not strongly connected graph (lian through which matter)

To use the stack

example: