poj 1236 Network of Schools

强连通分量题，涉及求强连通分量和缩点，基本是抄的模板，我也是把这题的代码当模板放这里。

做这题的一个收获，缩点后为构造强连通分量，最少要添加几条有向边：统计入度和出度为0的点的数量，取其大者。

#include <stdio.h>
#include <string.h>
#include <vector>
using namespace std;
const int N = 105;
int stk[N],col[N],dep[N],low[N];
int cnt,cols,top,ans1,ans2;
bool inq[N];
vector <int> a[N];
void DFS(int u)
{
    int i,t,l,v;
    l = a[u].size();
    dep[u] = low[u] = ++cnt;
    stk[top++] = u;
    inq[u] = 1;
    for(i = 0; i < l; i++)
    {
        v = a[u][i];
        if(!dep[v])
        {
            DFS(v);
            if(low[v] < low[u])
                low[u] = low[v];
        }
        else if(inq[v])
        {
            if(dep[v] < low[u])
                low[u] = dep[v];
        }
    }
    if(dep[u] == low[u])
    {
        cols++;
        do{
            t = stk[--top];
            inq[t] = 0;
            col[t] = cols;
        }while(t != u);
    }
}
void Tarjan(int n)
{
    int in0=0,out0=0,i,j,l,v;
    int in[N],out[N];
    cnt = top = cols = 0;
    memset(inq,0,sizeof inq);
    memset(dep,0,sizeof dep);
    memset(low,0,sizeof low);
    for(i = 1; i <= n; i++)
        if(!dep[i]) DFS(i);
    if(cols == 1)
    {
        ans1 = 1;
        ans2 = 0;
        return ;
    }
    memset(in,0,sizeof in);
    memset(out,0,sizeof out);
    for(i = 1; i <= n; i++)
    {
        l = a[i].size();
        for(j = 0; j < l; j++)
        {
            v = a[i][j];
            if(col[i] != col[v])
                out[col[i]]++, in[col[v]]++;
        }
    }
    for(i = 1; i <= cols; i++)
    {
        if(!in[i]) in0++;
        if(!out[i]) out0++;
    }
    ans1 = in0;
    ans2 = in0 > out0 ?in0 :out0 ;
}
int main()
{
    int n,i,t;
    while(~scanf("%d",&n))
    {
        for(i = 1; i <= n; i++)
        while(scanf("%d",&t),t)
            a[i].push_back(t);
        Tarjan(n);
        printf("%d\n%d\n",ans1,ans2);
    }
    return 0;
}