HDU 2767 Proving Equivalences——————Kosaraju求强连通分量

r囧r小猫 2022-01-30 20:45 217阅读 0赞

# [Proving Equivalences][] #

**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 ©, 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

--------------------

给你一个图，问你至少要加多少条边可以使这个图变成强连通的  
Koasraju求出DAG后  
num1 = DAG中入度为0点的个数  
num2 = DAG中出度为0点的个数

ans =max(num1, num2)

#include<bits/stdc++.h>
    using namespace std;
    const int MAXN = 1e5+7;
    int cmp[MAXN], ind[MAXN], outd[MAXN], V;
    bool used[MAXN];
    vector<int> G[MAXN];
    vector<int> rG[MAXN];
    vector<int> vs;
    
    void add_edge(int x,int y)
    {
        
            G[x].push_back(y);
            rG[y].push_back(x);
    }
    
    
    void dfs(int v)
    {
        
            used[v] = true;
            for(int i=0;i<(int)G[v].size();++i)
                    if(!used[G[v][i]])      dfs(G[v][i]);
            vs.push_back(v);
    }
    
    void rdfs(int v, int k)
    {
        
            used[v] = true;
            cmp[v]  = k;
            for(int i=0;i<(int)rG[v].size();++i)
                    if(!used[rG[v][i]])     rdfs(rG[v][i], k);
    }
    
    int scc()
    {
        
            memset(used, 0, sizeof(used));
            vs.clear();
            for(int v=0;v<V;++v)
                    if(!used[v])    dfs(v);
            memset(used, 0, sizeof(used));
            int k = 0;
            for(int i = vs.size() - 1; i >= 0; --i)
                    if(!used[vs[i]])        rdfs(vs[i], k++);
            return k;
    }
    
    int main()
    {
        
            // freopen("../in.txt","r",stdin);
            // freopen("../out.txt","w",stdout);
            int T, n, m;
            cin>>T;
            while(T--)
            {
        
                    cin>>n>>m;
                    V = n;
                    for(int i=0;i<=n;++i)
                    {
        
                            G[i].clear();
                            rG[i].clear();
                    }
                    for(int i = 0; i < m; ++i)
                    {
        
                            int x, y;
                            cin>>x>>y;
                            add_edge(x-1,y-1);
                    }
                    int N = scc();
                    if(N==1)
                    {
        
                            cout<<0<<endl;
                            continue;
                    }
                    memset(ind, 0, sizeof(ind));
                    memset(outd,0,sizeof(outd));
                    for(int v = 0; v < V; ++v)
                            for(int i = 0;i < (int)G[v].size(); ++i)
                                    if(cmp[v]!=cmp[G[v][i]])
                                    {
        
                                            ind[cmp[G[v][i]]]++;
                                            outd[cmp[v]]++;
                                    }
                    int num1 = 0;
                    int num2 = 0;
                    for(int i = 0; i < N; ++i)
                    {
        
                            if(!ind[i])      num1++;
                            if(!outd[i])     num2++;
                    }
                    cout<<max(num1, num2)<<endl;
            }
            return 0;
    }

[Proving Equivalences]: http://acm.hdu.edu.cn/showproblem.php?pid=2767