Proving Equivalences

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4410    Accepted Submission(s): 1562

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. 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
 

Source
NWERC 2008 

这个题居然是刘汝佳白皮书上第一个例题==！我就觉得眼熟嘛~

题意：把命题当做结点，推导当做游向彼岸，则本题就是给出n个结点m条边的有向图，要求加尽量少的边，使原图强连通

首先找出强连通分量，然后把每个强连通分量缩成一个点，得到一个DAG。接下来，设有n个结点（n个强连通分量）的入度为0，b个节点的出度为0 ，a b的最大值就是答案

/***************
hdu2762
2015.11.10
STL版tarjan
374MS 4104K 1911 B
***************/
#include <iostream>
#include<cstdio>
#include<stack>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 20005
vector<int>G[maxn];
int pre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt;
stack<int>S;
int in0[maxn],out0[maxn];
void dfs(int u)
{
    pre[u]=lowlink[u]=++dfs_clock;
    S.push(u);<span style="font-family: Arial, Helvetica, sans-serif;">//每搜索到一个点，压入栈中</span>
    for(int i=0;i<G[u].size();i++)//遍历与p相连的点
    {
        int v=G[u][i];
        if(!pre[v])//不在栈中
        {
            dfs(v);
            lowlink[u]=min(lowlink[u],lowlink[v]);
        }
        else if(!sccno[v])//在栈中
            lowlink[u]=min(lowlink[u],pre[v]);
    }
    if(lowlink[u]==pre[u])//发现一个根
    {
        scc_cnt++;
        for(;;)
        {
            int x=S.top();S.pop();//词典以上的所有点全部出栈 构成一个强连通分量
            sccno[x]=scc_cnt;//scc_cnt是强连通分量的序号
            if(x==u) break;
        }
    }
}
void find_scc(int n)
{
    dfs_clock=scc_cnt=0;
    memset(sccno,0,sizeof(sccno));
    memset(pre,0,sizeof(pre));
    for(int i=0;i<n;i++) if(!pre[i]) dfs(i);
}
int main()
{
    int  T,n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++) G[i].clear();
        for(int i=0;i<m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);u--;v--;
            G[u].push_back(v);
        }
        find_scc(n);
        for(int i=1;i<=scc_cnt;i++) in0[i]=out0[i]=1;
        for(int u=0;u<n;u++)
        {
            for(int i=0;i<G[u].size();i++)
            {
                int v=G[u][i];
                if(sccno[u]!=sccno[v]) in0[sccno[v]]=out0[sccno[u]]=0;
            }
        }
        int a=0,b=0;
        for(int i=1;i<=scc_cnt;i++)
        {
            if(in0[i]) a++;
            if(out0[i]) b++;
        }
        int ans=max(a,b);
        if(scc_cnt==1) ans=0;
        printf("%d\n",ans);
    }
    return 0;
}