試題編號(hào)：	201509-4
試題名稱：	高速公路
時(shí)間限制：	1.0s
內(nèi)存限制：	256.0MB
問題描述：	問題描述　　某國(guó)有n個(gè)城市，為了使得城市間的交通更便利，該國(guó)國(guó)王打算在城市之間修一些高速公路，由于經(jīng)費(fèi)限制，國(guó)王打算第一階段先在部分城市之間修一些單向的高速公路。　　現(xiàn)在，大臣們幫國(guó)王擬了一個(gè)修高速公路的計(jì)劃。看了計(jì)劃后，國(guó)王發(fā)現(xiàn)，有些城市之間可以通過高速公路直接（不經(jīng)過其他城市）或間接（經(jīng)過一個(gè)或多個(gè)其他城市）到達(dá)，而有的卻不能。如果城市A可以通過高速公路到達(dá)城市B，而且城市B也可以通過高速公路到達(dá)城市A，則這兩個(gè)城市被稱為便利城市對(duì)。　　國(guó)王想知道，在大臣們給他的計(jì)劃中，有多少個(gè)便利城市對(duì)。輸入格式　　輸入的第一行包含兩個(gè)整數(shù)n, m，分別表示城市和單向高速公路的數(shù)量。　　接下來m行，每行兩個(gè)整數(shù)a, b，表示城市a有一條單向的高速公路連向城市b。輸出格式　　輸出一行，包含一個(gè)整數(shù)，表示便利城市對(duì)的數(shù)量。樣例輸入5 51 22 33 44 23 5樣例輸出3樣例說明　　城市間的連接如圖所示。有3個(gè)便利城市對(duì)，它們分別是(2, 3), (2, 4), (3, 4)，請(qǐng)注意(2, 3)和(3, 2)看成同一個(gè)便利城市對(duì)。評(píng)測(cè)用例規(guī)模與約定　　前30%的評(píng)測(cè)用例滿足1 ≤ n ≤ 100, 1 ≤ m ≤ 1000；　　前60%的評(píng)測(cè)用例滿足1 ≤ n ≤ 1000, 1 ≤ m ≤ 10000；　　所有評(píng)測(cè)用例滿足1 ≤ n ≤ 10000, 1 ≤ m ≤ 100000。

問題鏈接：CCF201509試題。

問題描述：（參見上文）。

問題分析：這是一個(gè)強(qiáng)聯(lián)通圖的問題，用Tarjan算法來解決。另外一個(gè)算法是kosaraju算法，也用于解決強(qiáng)聯(lián)通圖問題。

程序說明：本程序采用Tarjan算法。主函數(shù)main()中，創(chuàng)建圖對(duì)象是參數(shù)本應(yīng)該用n，但是提交后出現(xiàn)了運(yùn)行錯(cuò)誤，所有改成n+1。程序通過使用Tarjan算法類（參見以下鏈接）來實(shí)現(xiàn)，做了簡(jiǎn)單修改，使用變量ans來存儲(chǔ)結(jié)果，其中增加了中間變量count。

求得強(qiáng)聯(lián)通子圖后，對(duì)于每一個(gè)強(qiáng)聯(lián)通子圖如果有k個(gè)結(jié)點(diǎn)，若k>1則強(qiáng)聯(lián)通對(duì)結(jié)點(diǎn)的數(shù)量為k*(k-1)/2，若k=1則為0。

相關(guān)鏈接：Tarjan算法查找強(qiáng)聯(lián)通組件。

提交后得100分的C++語言程序如下：

/* CCF201509-4 高速公路  */#include <iostream>#include <list>#include <stack>using namespace std;const int NIL = -1;int ans  = 0;// A class that rePResents an directed graphclass Graph{    int V;    // No. of vertices    list<int> *adj;    // A dynamic array of adjacency lists    // A Recursive DFS based function used by SCC()    void SCCUtil(int u, int disc[], int low[],                 stack<int> *st, bool stackMember[]);public:    Graph(int V);   // Constructor    void addEdge(int v, int w);   // function to add an edge to graph    void SCC();    // prints strongly connected components};Graph::Graph(int V){    this->V = V;    adj = new list<int>[V];}void Graph::addEdge(int v, int w){    adj[v].push_back(w);}// A recursive function that finds and prints strongly connected// components using DFS traversal// u --> The vertex to be visited next// disc[] --> Stores discovery times of visited vertices// low[] -- >> earliest visited vertex (the vertex with minimum//             discovery time) that can be reached from subtree//             rooted with current vertex// *st -- >> To store all the connected ancestors (could be part//           of SCC)// stackMember[] --> bit/index array for faster check whether//                  a node is in stackvoid Graph::SCCUtil(int u, int disc[], int low[], stack<int> *st,                    bool stackMember[]){    // A static variable is used for simplicity, we can avoid use    // of static variable by passing a pointer.    static int time = 0;    // Initialize discovery time and low value    disc[u] = low[u] = ++time;    st->push(u);    stackMember[u] = true;    // Go through all vertices adjacent to this    list<int>::iterator i;    for (i = adj[u].begin(); i != adj[u].end(); ++i)    {        int v = *i;  // v is current adjacent of 'u'        // If v is not visited yet, then recur for it        if (disc[v] == -1)        {            SCCUtil(v, disc, low, st, stackMember);            // Check if the subtree rooted with 'v' has a            // connection to one of the ancestors of 'u'            // Case 1 (per above discussion on Disc and Low value)            low[u]  = min(low[u], low[v]);        }        // Update low value of 'u' only of 'v' is still in stack        // (i.e. it's a back edge, not cross edge).        // Case 2 (per above discussion on Disc and Low value)        else if (stackMember[v] == true)            low[u]  = min(low[u], disc[v]);    }    // head node found, pop the stack and print an SCC    int w = 0;  // To store stack extracted vertices    int count = 0;    if (low[u] == disc[u])    {        while (st->top() != u)        {            w = (int) st->top();//            cout << w << " ";            count++;            stackMember[w] = false;            st->pop();        }        w = (int) st->top();//        cout << w << "/n";        count++;        stackMember[w] = false;        st->pop();    }    if(count > 1)        ans += count * (count -1) / 2;}// The function to do DFS traversal. It uses SCCUtil()void Graph::SCC(){    int *disc = new int[V];    int *low = new int[V];    bool *stackMember = new bool[V];    stack<int> *st = new stack<int>();    // Initialize disc and low, and stackMember arrays    for (int i = 0; i < V; i++)    {        disc[i] = NIL;        low[i] = NIL;        stackMember[i] = false;    }    // Call the recursive helper function to find strongly    // connected components in DFS tree with vertex 'i'    for (int i = 0; i < V; i++)        if (disc[i] == NIL)            SCCUtil(i, disc, low, st, stackMember);}int main(){    int n, m, src, dest;    cin >> n >> m;    Graph g(n+1);    for(int i=1; i<=m; i++) {        cin >> src >> dest;        g.addEdge(src, dest);    }    g.SCC();    cout << ans << endl;    return 0;}