强连通分量压缩是

 先缩点，然后计算各个强连通分量的入度为0的个数，出度为0的个数求他们最大值

#include <stdio.h>
#include <string.h>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
#define N 20005
stack<int>sta;
vector<int>mp[N];
int dfn[N];
int low[N];
int InStack[N];
int indexx,number;
int n, m;
int id[N];

void tarjan(int u)
{
    InStack[u] = 1;
    low[u] = dfn[u] = ++ indexx;
    sta.push(u);

    for (int i = 0; i < mp[u].size(); ++ i)
    {
        int t = mp[u][i];
        if (dfn[t] == 0)
        {
            tarjan(t);
            low[u] = min(low[u], low[t]);
        }
        else if (InStack[t] == 1)
        {
            low[u] = min(low[u], dfn[t]);
        }
    }

    if (low[u] == dfn[u])
    {
        ++ number;
        while (!sta.empty())
        {
            int v = sta.top();
            sta.pop();
            id[v]=number;
            InStack[v] = 0;
            if (v == u)
                break;
        }
    }
}

int main()
{
    //freopen("input.txt","r",stdin);
    int T;
    while(scanf("%d",&T)==1)
    {
        while(T--)
        {
            scanf("%d%d",&n,&m);
            memset(dfn, 0, sizeof(dfn));
            memset(low, 0, sizeof(low));
            memset(InStack, 0, sizeof(InStack));
            indexx = number = 0;
            for (int i = 1; i <= n; ++ i)
            {
                mp[i].clear();
            }
            while(!sta.empty())
                sta.pop();
            for(int i=1; i<=m; i++)
            {
                int a,b;
                scanf("%d%d",&a,&b);
                mp[a].push_back(b);
            }
            for(int i=1; i<=n; i++)
                if(!dfn[i])
                    tarjan(i);
            if(number==1)
            {
                printf("0\n");
                continue;
            }
            memset(dfn, 0, sizeof( dfn));
            memset(low, 0, sizeof (low));
            for(int i=1; i<=n; i++)//计算出入度
            {
                for(int j=0; j<mp[i].size(); j++)
                {
                    if(id[i]!=id[mp[i][j]]) dfn[id[i]]++, low[id[mp[i][j]]]++;//如果他们不在同一个的强连通分量话，把id【i】的出度++，id【他的节点】所在的连通分量的入度加加
                }
            }
            /*for(int i=1; i<=number; i++){
             printf("%d %d\n",dfn[i],low[i]);
            }*/
            int ans1=0, ans2=0;
            for(int i=1; i<=number; i++)
            {
                if(dfn[i]==0) ans1++;
                if(low[i]==0) ans2++;
            }
            printf("%d\n", max(ans1, ans2));
        }
    }
}

hdu 2767Proving Equivalences（强连通分量压缩 ）