分解强连通分量

模板题目：http://poj.org/problem?id=2553

Kosaraju：按顺序对原图和反图DFS一遍，在第二次DFS时点u可达未DFS的点v则u，v属于同一个联通分量

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<stack>

using namespace std;

const int maxn = 5005;
int out[maxn];
int in[maxn];
vector<int> G[maxn];
vector<int> Gr[maxn];
vector<int> rG[maxn];
vector<int> P[maxn];
vector<int> ps;
int ID[maxn];
int n, m;
bool vis[maxn];

void DFS(int u)
{
    vis[u] = true;
    for(int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        if(!vis[v]) DFS(v);
    }
    ps.push_back(u);
}

void RDFS(int u, int x)
{
    vis[u] = true;
    ID[u] = x;
    P[x].push_back(u);
    for(int i = 0; i < Gr[u].size(); i++)
    {
        int v = Gr[u][i];
        if(!vis[v]) RDFS(v, x);
    }
}

void init()
{
    memset(out, 0, sizeof(out));
    ps.clear();
    for(int i = 1; i <= n; i++)
    {
        G[i].clear();
        Gr[i].clear();
        rG[i].clear();
        P[i].clear();
    }
}

int Kosaraju()
{
    int res = 0;

    memset(vis, 0, sizeof(vis));
    for(int i =  1; i <= n; i++)
    {
        if(!vis[i]) DFS(i);
    }
    memset(vis, 0, sizeof(vis));
    for(int i = ps.size() - 1; i + 1; i--)
    {
        int v = ps[i];
        if(!vis[v]) RDFS(v, ++res);
    }
    return res;
}

void Get_DAG()
{
    for(int u = 1; u <= n; u++)
    {
        for(int i = 0; i < G[u].size(); i++)
        {
            int v = G[u][i];
            if(ID[u] != ID[v])
            {
                rG[ID[u]].push_back(ID[v]);
                out[ID[u]]++;
            }
        }
    }
}

int main()
{
    while(cin >> n, n)
    {
        cin >> m;
        init();
        for(int i = 0; i < m; i++)
        {
            int u, v;
            scanf("%d %d", &u, &v);
            G[u].push_back(v);
            Gr[v].push_back(u);
        }
        Kosaraju();
        Get_DAG();
        vector<int> ans;
        for(int i = 1; i <= n; i++)
        {
            if(!out[i] && P[i].size())
            {
                for(int j = 0; j < P[i].size(); j++)
                {
                    int v = P[i][j];
                    ans.push_back(v);
                }
            }
        }
        sort(ans.begin(), ans.end());
        for(int i = 0; i < ans.size(); i++)
        {
            int v = ans[i];
            cout << v << (v == ans.back() ? '\n' : ' ');
        }
    }
    return 0;
}

Tarjan：对于一个强连通分量，其中的第一个点被dfs搜到的时候，接着一定是整个联通分量被遍历完并且回到最开始的点，这时候就可以对该dfs子树进行整体编号

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<stack>

using namespace std;

const int maxn = 5005;
int out[maxn];
stack<int> S;
vector<int> G[maxn];
vector<int> rG[maxn];
vector<int> P[maxn];
bool isinstack[maxn];
int low[maxn], dfn[maxn];
int ID[maxn];
int n, m, ssc_num, index;


void Get_DAG()
{
    for(int u = 1; u <= n; u++)
    {
        for(int i = 0; i < G[u].size(); i++)
        {
            int v = G[u][i];
            if(ID[u] != ID[v])
            {
                rG[ID[u]].push_back(ID[v]);
                out[ID[u]]++;
            }
        }
    }
}

void init()
{
    memset(out, 0, sizeof(out));
    memset(dfn, -1, sizeof(dfn));
    memset(isinstack, 0, sizeof(isinstack));
    index = 1;
    for(int i = 1; i <= n; i++)
       P[i].clear(), G[i].clear(), rG[i].clear();
}

void TarDFS(int u)
{
    dfn[u] = index++;
    low[u] = dfn[u];
    isinstack[u] = 1;
    S.push(u);
    for(int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        if(dfn[v] == -1)
        {
            TarDFS(v);
            low[u] = min(low[u], low[v]);
        }
        else if(isinstack[v])
        {
            low[u] = min(low[u], dfn[v]);
        }
    }

    if(dfn[u] ==low[u])
    {
        ssc_num++;
        while(S.top() != u)
        {
            isinstack[S.top()] = 0;
            ID[S.top()] = ssc_num;
            P[ssc_num].push_back(S.top());
            S.pop();
        }
        isinstack[u] = 0;
        ID[u] = ssc_num;
        P[ssc_num].push_back(u);
        S.pop();
    }
}

int Tarjan()
{
    ssc_num = 0;
    index = 1;
    for(int i = 1; i <= n; i++)
    {
        if(dfn[i] == -1)
        {
            TarDFS(i);
        }
    }
    return ssc_num;
}

int main()
{
    while(cin >> n, n)
    {
        cin >> m;
        init();
        for(int i = 0; i < m; i++)
        {
            int u, v;
            scanf("%d %d", &u, &v);
            G[u].push_back(v);
        }
        Tarjan();
        Get_DAG();
        vector<int> ans;
        for(int i = 1; i <= n; i++)
        {
            if(!out[i] && P[i].size())
            {
                for(int j = 0; j < P[i].size(); j++)
                {
                    int v = P[i][j];
                    ans.push_back(v);
                }
            }
        }
        sort(ans.begin(), ans.end());
        for(int i = 0; i < ans.size(); i++)
        {
            int v = ans[i];
            cout << v << (v == ans.back() ? '\n' : ' ');
        }
    }
    return 0;
}