ACM模版

Tarjan

/* * Tarjan算法 * 复杂度O(N+M) */
const int MAXN = 20010; // 点数
const int MAXM = 50010; // 边数

struct Edge
{
    int to, next;
}edge[MAXM];

int head[MAXN], tot;
int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN];    // Belong数组的值是1~scc
int Index, top;
int scc;                                                // 强连通分量的个数
bool Instack[MAXN];
int num[MAXN];                                          // 各个强连通分量包含点的个数,数组编号1~scc
                                                        // num数组不一定需要,结合实际情况
void addedge(int u, int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
    return ;
}

void Tarjan(int u)
{
    int v;
    Low[u] = DFN[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        v = edge[i].to;
        if (!DFN[v])
        {
            Tarjan(v);
            if (Low[u] > Low[v])
            {
                Low[u] = Low[v];
            }
        }
        else if (Instack[v] && Low[u] > DFN[v])
        {
            Low[u] = DFN[v];
        }
    }
    if (Low[u] == DFN[u])
    {
        scc++;
        do
        {
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = scc; num[scc]++;
        }
        while (v != u);
    }
    return ;
}

void solve(int N)
{
    memset(DFN, 0, sizeof(DFN));
    memset(Instack, false, sizeof(Instack));
    memset(num, 0, sizeof(num));
    Index = scc = top = 0;
    for (int i = 1; i <= N; i++)
    {
        if (!DFN[i])
        {
            Tarjan(i);
        }
    }
    return ;
}

void init()
{
    tot = 0;
    memset(head, -1, sizeof(head));
    return ;
}

Kosaraju

/* * 复杂度O(N+M) */
const int MAXN = 20010;
const int MAXM = 50010;

struct Edge
{
    int to, next;
} edge1[MAXM], edge2[MAXM];     // edge1是原图G,edge2是逆图GT

int head1[MAXN], head2[MAXN];
bool mark1[MAXN], mark2[MAXN];
int tot1, tot2;
int cnt1, cnt2;
int st[MAXN];       // 对原图进行dfs,点的结束时间从小到大排序
int Belong[MAXN];   // 每个点属于哪个连通分量(0~cnt2-1)
int num;            // 中间变量,用来数某个连通分量中点的个数
int setNum[MAXN];   // 强连通分量中点的个数,编号0~cnt2-1

void addedge(int u, int v)
{
    edge1[tot1].to = v;
    edge1[tot1].next = head1[u];
    head1[u] = tot1++;
    edge2[tot2].to = u;
    edge2[tot2].next = head2[v];
    head2[v] = tot2++;
    return ;
}

void DFS1(int u)
{
    mark1[u] = true;
    for (int i = head1[u]; i != -1; i = edge1[i].next)
    {
        if(!mark1[edge1[i].to])
        {
            DFS1(edge1[i].to);
        }
    }
    st[cnt1++] = u;
    return ;
}

void DFS2(int u)
{
    mark2[u] = true;
    num++;
    Belong[u] = cnt2;
    for (int i = head2[u]; i != -1; i = edge2[i].next)
    {
        if(!mark2[edge2[i].to])
        {
            DFS2(edge2[i].to);
        }
    }
    return ;
}

void solve(int n)   // 点的编号从1开始
{
    memset(mark1, false, sizeof(mark1));
    memset(mark2, false, sizeof(mark2));
    cnt1 = cnt2 = 0;
    for (int i = 1; i <= n; i++)
    {
        if (!mark1[i])
        {
            DFS1(i);
        }
    }
    for (int i = cnt1 - 1; i >= 0; i--)
    {
        if (!mark2[st[i]])
        {
            num = 0;
            DFS2(st[i]);
            setNum[cnt2++] = num;
        }
    }
    return ;
}