HDU2767-Proving Equivalences(强连通+缩点+DGA性质)--Tarjan模板

Consider the following exercise, found in a generic linear algebra textbook.

Let A be an n × n matrix. Prove that the following statements are equivalent:

1. A is invertible.
2. Ax = b has exactly one solution for every n × 1 matrix b.
3. Ax = b is consistent for every n × 1 matrix b.
4. Ax = 0 has only the trivial solution x = 0.

The typical way to solve such an exercise is to show a series of implications. For instance, one can proceed by showing that (a) implies (b), that (b) implies (c), that (c) implies (d), and finally that (d) implies (a). These four implications show that the four statements are equivalent.

Another way would be to show that (a) is equivalent to (b) (by proving that (a) implies (b) and that (b) implies (a)), that (b) is equivalent to (c), and that (c) is equivalent to (d). However, this way requires proving six implications, which is clearly a lot more work than just proving four implications!

I have been given some similar tasks, and have already started proving some implications. Now I wonder, how many more implications do I have to prove? Can you help me determine this?
Input
On the first line one positive number: the number of testcases, at most 100. After that per testcase:

• One line containing two integers n (1 ≤ n ≤ 20000) and m (0 ≤ m ≤ 50000): the number of statements and the number of implications that have already been proved.

• m lines with two integers s1 and s2 (1 ≤ s1, s2 ≤ n and s1 ≠ s2) each, indicating that it has been proved that statement s1 implies statement s2.
  Output
  Per testcase:

• One line with the minimum number of additional implications that need to be proved in order to prove that all statements are equivalent.
  Sample Input
  2
  4 0
  3 2
  1 2
  1 3
  Sample Output
  4
  2
  题目：HDU2767
  题意：给你一些语句编号1~n，其中有些语句Xi能够证明Xj，问（最少）还需要证明几条语句之间的关系，使得所有语句之间的关系成为一个环。
  思路：题意说的很明确了，问最少加几条边能够使一个DGA（有向无环图）成为一个环，但是题目所给的图中可能含有环，所以我们先用Tarjan算法缩点，得到一个DGA，然后利用DGA的性质：将一个DGA变成一个环，最少需要加的边数=max（出度为0的点的个数，入度为0的点的个数），即可得到答案。
  AC代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define met(s,k) memset(s,k,sizeof s)
#define scan(a) scanf("%d",&a)
#define scanl(a) scanf("%lld",&a)
#define scann(a,b) scanf("%d%d",&a,&b)
#define scannl(a,b) scanf("%lld%lld",&a,&b)
#define scannn(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define prin(a) printf("%d\n",a)
#define prinl(a) printf("%lld\n",a)
using namespace std;
typedef long long ll;
const int maxn=20005;
const double eps=1e-4;
int dfn[maxn],dep[maxn],c[maxn],r[maxn],color[maxn],p[maxn],num,stacks[maxn],vis[maxn],cont,lid,k,n,m;
struct edge
{
    int en,next;
}e[3*maxn];
void init()
{
    met(p,-1);
    num=0;
    k=0;
    cont=1;
    lid=-1;
    met(c,0);
    met(r,0);
    met(vis,0);
    met(dfn,0);
    met(dep,0);
    met(color,0);
}
void add(int st,int en)
{
    e[num].en=en;
    e[num].next=p[st];
    p[st]=num++;
}
void tarjan(int u)
{
    vis[u]=1;
    dfn[u]=dep[u]=cont++;
    stacks[++lid]=u;
    for(int i=p[u];i!=-1;i=e[i].next)
    {
        int v=e[i].en;
        if(vis[v]==0)tarjan(v);
        if(vis[v]==1)dep[u]=min(dep[u],dep[v]);
    }
    if(dfn[u]==dep[u])
    {
        k++;
        do
        {
            color[stacks[lid]]=k;
            vis[stacks[lid]]=-1;
        }
        while(stacks[lid--]!=u);
    }
}
int main()
{
    int t;
    scan(t);
    while(t--)
    {
        scann(n,m);
        init();
        for(int i=0;i<m;i++)
        {
            int x,y;
            scann(x,y);
            add(x,y);
        }
        for(int i=1;i<=n;i++)
        {
            if(vis[i]==0)tarjan(i);
        }
        if(k==1)
        {
            prin(0);
            continue;
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=p[i];j!=-1;j=e[j].next)
            {
                  int v=e[j].en;
                  if(color[i]!=color[v])
                  {
                      c[color[i]]++;
                      r[color[v]]++;
                  }
            }
        }
        int chu=0,ru=0;
        for(int i=1;i<=k;i++)
        {
            if(c[i]==0)chu++;
            if(r[i]==0)ru++;
        }
        prin(max(chu,ru));
    }
    return 0;
}