HDU2767

/*
*题目大意：
*        在数学里面有两种关系，一种是充分条件，即对于集合p,q，p => q，
*        另一种是等价关系，p => q && q =>p, 这两种关系都具有传递性，p
*        => q 可以对应到有节点p到节点q有一条边。问：给定一些集合的充分
*        性关系，确定出若让所有集合都等价，还需在添加最少的充分性条件。
*解题思路：
*        有向图加最少边变强连通，模板题
*/

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int MAXN = 20005;

typedef struct _node
{
    int v, next;
}N;
N edge[MAXN * 3];
int dfn[MAXN], low[MAXN], step;
int inS[MAXN], id[MAXN], scc, myS[MAXN], top;
int in[MAXN], out[MAXN], cntEdge, head[MAXN];

void init()
{
    cntEdge = step = scc = top = 0;
    for(int i = 0; i < MAXN; i++)
    {
        head[i] = -1;
        dfn[i] = low[i] = -1;
        id[i] = -1;
        in[i] = out[i] = 0;
        inS[i] = 0;
    }
}

void tarjan(int n)
{
    dfn[n] = low[n] = ++step;
    myS[top++] = n;
    inS[n] = 1;
    for(int f = head[n]; f != -1; f = edge[f].next)
    {
        int son = edge[f].v;
        if(dfn[son] == -1)
        {
            tarjan(son);
            low[n] = min(low[n], low[son]);
        }
        else if(inS[son] != 0)
            low[n] = min(low[n], dfn[son]);
    }

    if(low[n] == dfn[n])
    {
        int tmp;
        do
        {
            tmp = myS[--top];
            inS[tmp] = 0;
            id[tmp] = scc;
        }while(myS[top] != n);
        scc++;
    }
}

void addEdge(int u, int v)
{
    edge[cntEdge].v = v;
    edge[cntEdge].next = head[u];
    head[u] = cntEdge++;
}



int main(void)
{
#ifndef ONLINE_JUDGE 
    freopen("inHDU2767.txt", "r", stdin);
#endif

    int n, m, cas;
    scanf("%d", &cas);
    while(cas--)
    {
        scanf("%d %d", &n, &m);
        init();
        int u, v;
        for(int i = 0; i < m; i++)
        {
            scanf("%d %d", &u, &v);
            addEdge(u, v);
        }

        for(int i = 1; i <= n; i++)
        {
            if(dfn[i] == -1)
                tarjan(i);
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = head[i]; j != -1; j = edge[j].next)
            {
                u = i, v = edge[j].v;
                if(id[u] == id[v])
                    continue;
                else
                {
                    in[id[v]]++;
                    out[id[u]]++;
                }
            }    
        }
        int inNum = 0, outNum = 0;
        for(int i = 0; i < scc; i++)
        {
            if(!in[i])
                inNum++;
            if(!out[i])
                outNum++;
        }
        if(scc >= 2)
            printf("%d\n", max(inNum, outNum));
        else
            printf("0\n");
    }
    return 0;
}