poj2553

题意：求出度为0的强连通分支。

分析：采用邻接阵的强连通分支tarjan模板。

const int maxn = 5005, maxm = 100005;

struct Edge
{
    int v, next;
}edge[maxm];

int n, m, ncount, idx;
int head[maxn];
int dfn[maxn], low[maxn];
bool vis[maxn], instk[maxn];
int stk[maxn], top, cpn_num, cpn[maxn];
int out[maxn];

void addedge(int a, int b)
{
    //printf("%d %d\n", a, b);
    edge[ncount].v = b;
    edge[ncount].next = head[a];
    head[a] = ncount++;
}

void tarjan(int u)
{
    vis[u] = true;
    stk[top++] = u;
    instk[u] = true;
    dfn[u] = low[u] = idx++;
    for (int i = head[u]; ~i; i = edge[i].next)
    {
        int v = edge[i].v;
        if (!vis[v])
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }else if (instk[v])
            low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] != low[u])
        return;
    int v;
    do
    {
        v = stk[--top];
        instk[v] = false;
        cpn[v] = cpn_num;
    }while(u != v);
    cpn_num++;
}

CALL:
        memset(head, -1, sizeof(head));
        memset(vis, 0, sizeof(vis));
        memset(instk, 0, sizeof(instk));
        memset(out, 0, sizeof(out));
        input();
        idx = 0;
        top = 0;
        cpn_num = 0;
        for (int i = 0; i < n; i++)
            if (!vis[i])
                tarjan(i);