题目链接

题意： 给定一张有向图。找出全部强连通分量，并输出。

思路：有向图的强连通分量用Tarjan算法，然后用map映射，便于输出，注意输出格式。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <map>

#include <algorithm>

using namespace std;

const int MAXN = 2000;

const int MAXM = 50000;

struct Edge{

    int to, next;

}edge[MAXM];

int head[MAXN], tot;

int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN];

int Index, top;

int scc;

bool Instack[MAXN];

int num[MAXN];

int n, m, cnt;

map<string, int> sTon;

map<int, string> nTos; 

void init() {

    tot = cnt = 0;

    memset(head, -1, sizeof(head));

    sTon.clear();

    nTos.clear();

}

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() {

    memset(Low, 0, sizeof(Low));

    memset(DFN, 0, sizeof(DFN));

    memset(num, 0, sizeof(num));

    memset(Belong, 0, sizeof(Belong));

    memset(Stack, 0, sizeof(Stack));

    memset(Instack, false, sizeof(Instack));

    Index = scc = top = 0;

    for (int i = 1; i <= n; i++)

        if (!DFN[i])

            Tarjan(i);

}

int main() {

    int t = 0;

    while (scanf("%d%d", &n, &m)) {

        if (n == 0 && m == 0) break;

        init();

        string s1, s2;

        for (int i = 0; i < m; i++) {

            cin >> s1 >> s2;

            if (sTon.find(s1) == sTon.end()) {

                cnt++;

                sTon[s1] = cnt;

                nTos[cnt] = s1;

            }

            if (sTon.find(s2) == sTon.end()) {

                cnt++;

                sTon[s2] = cnt;

                nTos[cnt] = s2;

            }

            addedge(sTon[s1], sTon[s2]);

        }

        if (t) printf("\n");

        printf("Calling circles for data set %d:\n", ++t);

        solve();

        for (int i = 1; i <= scc; i++) {

            int flag = 0;

            for (int u = 1; u <= n; u++) {

                if (Belong[u] == i) {

                    if (!flag) {

                        cout << nTos[u];

                        flag = 1;

                    }

                    else cout << ", " << nTos[u];

                }

            }

            printf("\n");

        }

    }

    return 0;

}