求无向图的割顶和桥【模板】

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

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

int n, m, tot, dfs_clock;

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

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;//注意这里一定是++dfs_clock,不然后来判断新搜到的节点有没有访问过就会出错。
    int child = 0;
    for (int i = h[u]; ~i; i = a[i].next)
    {
        int v = a[i].v;
        if (!dfn[v])//如果这个节点我还没有访问过，树边
        {
            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);//桥和割点的判断略有不同
            }
        }else if (dfn[v] < dfn[u] && v != fa)//反向边
        {
            lowu = min(lowu, dfn[v]);
        }
    }
    if (fa == 0 && child == 1)//排除根节点的特殊情况。
    {
        iscut[u] = 0;
    }
    low[u] = lowu;
    return lowu;
}

int main()
{
    freopen("无向图的割顶和桥.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 (iscut[i])
            printf("%d ", i);
    return 0;
}

现在力求代码规范，模板不错，所以这是照着白书打的，以前打的怎样都忘了就好。