【模板】tarjan（强连通+缩点）

具体学习参考https://blog.youkuaiyun.com/qq_34374664/article/details/77488976

/*
* Tarjan 算法
* 复杂度 O(N+M)
*/
#include<bits/stdc++.h>
using namespace std;
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++;
}
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);
    }
}
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);
}
void init()
{
    tot = 0;
    memset(head, - 1,sizeof(head));
}