tarjan算法----求强连通分量

扔上带注释的代码 回去慢慢补==

#include
#include
#include
#include
#include
#define max_edge 10010
#define max_node 10010
using namespace std;

struct Edge {
    int u;
    int v;
    Edge* next;
} edge[max_edge];

Edge* v[max_node]; //节点i的第一条边 
int DFN[max_node], LOW[max_node]; // DFN[i]为i的时间戳 LOW[i]为i能回溯到的最高级祖先 
bool instack[max_node]; //判断是否在栈内 
int Stap[max_node], Stop; //定义栈 
int Dindex; //标号 时间戳 
int Bcnt, Belong[max_node]; //bcnt为强连通块数量，belong[i]表示i属于哪个强连通块 
int n; //总节点数 

void tarjan(int i) {
    int j;
    DFN[i] = LOW[i] = ++Dindex;
    instack[i] = true;
    Stap[++Stop] = i;
    for(Edge *e = v[i]; e; e = e->next) {
        j = e->v;
        if(!DFN[j]) {
            tarjan(j);
            LOW[i] = min(LOW[i], LOW[j]);
        } else if(instack[j] && DFN[j] < LOW[i])
            LOW[i] = DFN[j];
    }
    if(DFN[i] == LOW[i]) {
        Bcnt++;
        do {
            j = Stap[Stop--];
            instack[j] = false;
            Belong[j] = Bcnt;
        }
        while (j!=i);
    }
}

void solve() {
    int i;
    Stop = Bcnt = Dindex = 0;
    memset(DFN, 0, sizeof(DFN));
    for(int i = 1; i <= n; i++)
        if(!DFN[i]) tarjan(i);
}

int main() {
    int m;
    cin >> n >> m; 
    for(int i = 1; i <= m; i++) {
        cin >> edge[i].u >> edge[i].v;
        edge[i].next = v[edge[i].u];
        v[edge[i].u] = &edge[i];
    }
    solve();
    return 0;
}

byvoid的详解