HDU2767 Proving Equivalences

Proving Equivalences

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

Total Submission(s): 3027 Accepted Submission(s): 1133

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

题意：要完成n个命题的等价性证明，以进行了m次，问最少还要几次。

应该算一个强连通的模板题，把每个强连通分量看成一个点，再看出度为0，入度为0中这两个钟最大的就是答案。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
using namespace std;
const int MAXN=20010;
vector<int> G[MAXN];
int DFN[MAXN],LOW[MAXN],sccno[MAXN],dfs_clock,scc_cnt;
stack<int> S;
void dfs(int u)
{
    DFN[u]=LOW[u]=++dfs_clock;  //图的每个节点标号
    S.push(u);  //每搜到一个节点，就入栈
    for(int i=0;i<G[u].size();i++)
    {
        int v=G[u][i];
        if(!DFN[v]) //u和v相连，v不在栈中，所以u的low值为u和v中小的一个
        {
            dfs(v);
            LOW[u]=min(LOW[u],LOW[v]);
        }
        else if(!sccno[v])  //注意这里是在栈中，sccno不为0就说明已经出栈了
        {
            LOW[u]=min(LOW[u],DFN[v]);
        }
    }
    if(LOW[u]==DFN[u])  //u是强连通分量
    {
        scc_cnt++;
        while(1)
        {
            int x=S.top();  //退栈
            S.pop();
            sccno[x]=scc_cnt;
            if(x==u)
                break;
        }
    }
}
int in[MAXN],out[MAXN];
int main()
{
    int t,n,m,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
            G[i].clear();
        int u,v;
        while(m--)
        {
            scanf("%d%d",&u,&v);
            G[u].push_back(v);
        }
        dfs_clock=scc_cnt=0;
        memset(DFN,0,sizeof(DFN));
        memset(sccno,0,sizeof(sccno));
        for(i=1;i<=n;i++)
            if(!DFN[i])
            dfs(i);
        if(scc_cnt==1)
        {
            printf("0\n");
            continue;
        }
        for(i=1;i<=scc_cnt;i++)
        {
            in[i]=out[i]=1;
        }
        for(u=1;u<=n;u++)
        {
            for(i=0;i<G[u].size();i++)
            {
                v=G[u][i];
                if(sccno[u]!=sccno[v])
                    in[sccno[v]]=out[sccno[u]]=0;
            }
        }
        int a=0,b=0;
        for(i=1;i<=scc_cnt;i++)
        {
            if(in[i])
                a++;
            if(out[i])
                b++;
        }
        int ans=max(a,b);
        printf("%d\n",ans);
    }
    return 0;
}