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。

View Code

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);