HDU 2767 Proving Equivalences【Tarjan算法+缩点】

传送门：HDU 2767

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Proving Equivalences Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Problem Description
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.A x = b has exactly one solution for every n × 1 matrix b.
3.A x = b is consistent for every n × 1 matrix b.
4.A x = 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 ©, that © 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 ©, and that © 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

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题意：

给定一些已经存在的等价性证明，要求全部等价，需要在多最少几次证明。
（给出n个点和m条边，问最少须添加多少个边，使得任意两点间能互通。给出的边是有向的。）

题解：

这题目我搞了一天，几乎就是花了一天时间几乎看懂Tarjan算法这个算法挺难搞 。关于图的强连通性我之前只会Kosaraju算法，不过这个算法时间复杂度太高了，这题做不了，我哭。

其实就是问你加几条边使原来的图变成强连通图。先求出强连通分量，然后进行缩点，在缩点后的图上统计入度和出度为0结点的最大值，就是需要加的边数，注意如果整个图已经是强连通，就直接是答案。
建议先去熟悉tarjan算法

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AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<stack>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int N=250005;
int t;
int n,m;
int dfn[N],low[N];
int belong[N],head[N];
int bcnt,k,idx;//bcnt为强连通分量的个数
int vis[N];//判断点是否在栈内
int in[N],out[N];
struct edge
{
  int to,next;
}e[N*2];
stack<int> s;
void add(int u,int v)//建边
{
  e[k].to=v;
  e[k].next=head[u];
  head[u]=k++;
}
void tarjan(int u)
{
  int v;
  dfn[u]=low[u]=++idx;
  vis[u]=1;
  s.push(u);
  for(int i=head[u];i!=-1;i=e[i].next)
  {
    v=e[i].to;
    if(!dfn[v])
    {
      tarjan(v);
      low[u]=min(low[u],low[v]);
    }
    else if(vis[v])
    {
      low[u]=min(low[u],dfn[v]);
    }
  }
  if(dfn[u]==low[u])
  {
    bcnt++;
    do
    {
      v=s.top();
      s.pop();
      vis[v]=0;
      belong[v]=bcnt;
    }while(u!=v);
  }
}
int main()
{
  int a,b;
  scanf("%d",&t);
  while(t--)
  {
    //初始化
    while(!s.empty()) s.pop();
    memset(head,-1,sizeof(head));
    memset(vis,0,sizeof(vis));
    memset(belong,0,sizeof(belong));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    scanf("%d%d",&n,&m);
    bcnt=0,k=1,idx=0;
    for(int i=0;i<m;i++)
    {
      scanf("%d%d",&a,&b);
      add(a,b);
    }
    for(int i=1;i<=n;i++)
    {
      if(!dfn[i])
      {
        tarjan(i);
      }
    }
    if(bcnt==1)//已经是强连通分量
    {
      printf("0\n");
      continue;
    }
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
    int v;
    //计算缩点后的出度和入度
    for(int u=1;u<=n;u++)
    {
      for(int i=head[u];i!=-1;i=e[i].next)
      {
        v=e[i].to;
        if(belong[v]!=belong[u])
        {
          in[belong[v]]=out[belong[u]]=1;
        }
      }
    }
    int ansin=0,ansout=0;
    for(int i=1;i<=bcnt;i++)//求出缩点后入度和出度为0的强连通分量的个数
    {
      if(!in[i]) ansin++;
      if(!out[i]) ansout++;
    }
    printf("%d\n",max(ansin,ansout));
  }
  return 0;
}