强连通分量--tarjan

tarjan

算法思想

模板

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 10100;
const int M = 1e9+7;
int n,m,idx,top,cnt;

int dfn[maxn];      //这个点是第几个被访问的
int low[maxn];      //这个点所能到的点的dfn的最小值
bool vis[maxn];     //是否在s中
int s[maxn];        //栈

vector<int> a[maxn];        //图

void tarjan(int u)
{
    dfn[u] = low[u] = ++idx;
    s[++top] = u;
    vis[u] = 1;
    for(auto v : a[u])
    {
        if(!dfn[v])         //如果这个点还没被访问过
        {
            tarjan(v);
            low[u] = min(low[u],low[v]);    //回溯
        }
        else if(vis[v])     //如果这个点在栈内
        {
            low[u] = min(low[u],low[v]);
        }
    }
    if(dfn[u] == low[u])        //找到一组强连通分量
    {
        if(s[top] != u) cnt++;
        while (s[top] != u)
        {
            vis[s[top]] = 0;
            top--;
        }
        vis[s[top]] = 0;
        top--;
    }
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
#endif
    cin(n),cin(m);
    for(int i = 0,u,v; i < m; i++)
    {
        scanf("%d%d",&u,&v);
        a[u].push_back(v);
    }
    for(int i = 1; i <= n; i++)
    {
        if(dfn[i] == 0)
        {
            tarjan(i);
        }
    }
    cout<<cnt<<endl;
    return 0;
}

例题

https://www.luogu.com.cn/problem/P2863

参考博客

初探tarjan算法