强联通分量 Tarjan算法

//上交红书模板
// 图有节点0~n-1，共n个节点
vector< pair<int, int> > g; //存有向边
vector<int> ans;  //染色结果

struct strongly_connected_components{
    vector<int> &color;
    vector<int> Stack;
    int colorCnt, curr, *instack, *dfn, *low, *info, *next, *to;

    void dfs ( int x ) {
        dfn[x] = low[x] = ++ curr;
        Stack.push_back( x );
        instack[x] = true;
        for ( int j = info[x]; j; j = next[j] ){
            if ( ! instack[to[j]] ){
                dfs ( to[j] );
                low[x] = min ( low[x], low[to[j]] );
            }
            else{
                if ( instack[to[j]] == 1 ){
                    low[x] = min ( low[x], dfn[to[j]] );
                }
            }
        }
        if ( low[x] == dfn[x] ){
            while ( Stack.back() != x ){
                color[Stack.back()] = colorCnt;
                instack[Stack.back()] = 2;
                Stack.pop_back();
            }
            color[Stack.back()] = colorCnt++;
            instack[Stack.back()] = 2;
            Stack.pop_back();
        }
    }

    strongly_connected_components ( const vector< pair<int, int> > &edgeList, 
                            int n, vector<int> &ans ) : color(ans) {
        color.resize( n );
        instack = new int[n];
        dfn = new int[n];
        low = new int[n];
        info = new int[n];
        next = new int[(int)edgeList.size() + 5];
        to = new int[(int)edgeList.size() + 5];
        fill_n( info, n, 0 );
        for ( size_t i = 0; i < edgeList.size(); ++ i ){
            to[i+1] = edgeList[i].second;
            next[i+1] = info[edgeList[i].first];
            info[edgeList[i].first] = i + 1;
        }
        fill_n( instack, n, 0 );
        colorCnt = 0;
        curr = 0;
        for ( int i = 0; i < n; i ++ ){
            if ( ! instack[i] ){
                dfs ( i );
            }
        }
        delete[] instack;
        delete[] dfn;
        delete[] low;
        delete[] info;
        delete[] next;
        delete[] to;
    }
};

int main()
{
    int m, n;
    while ( cin >> m >> n ){
        int a, b;
        g.clear();
        ans.clear();
        for ( int i = 0; i < n; i ++ ){
            cin >> a >> b;
            a --;
            b --;
            pair<int, int> tmp;
            tmp.first = a;
            tmp.second = b;
            g.push_back ( tmp );
        }
        strongly_connected_components scc ( g, m, ans );
        for ( int i = 0; i < ans.size(); i ++ ){
            cout << ans[i] << ' ';
        }
    }
}

网上tarjan模板

HDOJ1269

图中任意两点强联通

#define MAX_V 11111
 
int V, E;
int dfn[MAX_V], low[MAX_V];
bool instack[MAX_V];
int din;
int cnt;
stack<int>s;
vector<int>Edge[MAX_V];
 
void tarjan(int x){
    instack[x] = true;
    dfn[x] = low[x] = din++;
    s.push(x);
    for (int i = 0; i < Edge[x].size(); i++){
        int j = Edge[x][i];
        if (!dfn[j]){
            tarjan(j);
            low[x] = min(low[x], low[j]);
        }
        else if (instack[j])
            low[x] = min(low[x], dfn[j]);
    }
    if (dfn[x] == low[x]){
        cnt++;
        int tmp;
        do{
            tmp = s.top();
            s.pop();
            instack[tmp] = false;
        }while (x != tmp);
    }
}

int main(){
    while (scanf("%d %d", &V, &E) && (V || E)){
        for (int i = 0; i < V; i++)
            Edge[i].clear();
        for (int i = 0; i < E; i++){
            int u, v;
            scanf("%d %d", &u, &v);
            Edge[u - 1].push_back(v - 1);
        }
        memset(dfn, 0, sizeof(dfn));
        memset(instack, false, sizeof(instack));
        din = 0;
        cnt = 0;
        for (int i = 0; i < V; i++){
            if(!dfn[i])
                tarjan(i);
        }
        if(cnt == 1)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}