hdu 2767(强连通分量)

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本文探讨了如何通过证明一系列等价性来确定多个陈述的等效性，包括矩阵运算、线性代数概念及证明技巧。通过实例分析，展示了简化证明过程的方法，以及在特定任务中所需额外证明步骤的数量。

Proving Equivalences

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 584 Accepted Submission(s): 238

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

Sample Output

4
2

Source

NWERC 2008

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题目： http://acm.hdu.edu.cn/showproblem.php?pid=2767

分析：模板题，可以用来测模板。。。注意缩成1个点的情况特殊判掉

代码：

#include<cstdio>
#define min(a,b) a<b?a:b
#define max(a,b) a>b?a:b
using namespace std;
const int mm=50005;
const int mn=20002;
int s[mm],t[mm],p[mm];
int id[mn],q[mn],dfn[mn],low[mn],h[mn];
int i,j,k,n,m,qe,tsp,cnt,cas,sum1,sum2;
void dfs(int u)
{
    int i,v;
    dfn[u]=low[u]=++tsp;
    q[qe++]=u;
    for(i=h[u];i>=0;i=p[i])
        if(!dfn[v=t[i]])
            dfs(v),low[u]=min(low[u],low[v]);
        else if(id[v]<0)low[u]=min(low[u],dfn[v]);
    if(low[u]==dfn[u])
    {
        id[u]=++cnt;
        while((v=q[--qe])!=u)id[v]=cnt;
    }
}
void tarjan()
{
    int i;
    for(qe=tsp=cnt=i=0;i<=n;++i)dfn[i]=0,id[i]=-1;
    for(i=1;i<=n;++i)
        if(!dfn[i])dfs(i);
}
int main()
{
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d%d",&n,&m);
        for(i=0;i<=n;++i)h[i]=-1;
        for(k=0;k<m;++k)
        {
            scanf("%d%d",&i,&j);
            s[k]=i,t[k]=j,p[k]=h[i],h[i]=k;
        }
        tarjan();
        if(cnt==1)
        {
            printf("0\n");
            continue;
        }
        for(i=1;i<=cnt;++i)p[i]=q[i]=0;
        for(k=0;k<m;++k)
            if((i=id[s[k]])!=(j=id[t[k]]))
                ++p[i],++q[j];
        for(sum1=sum2=0,i=1;i<=cnt;++i)
            sum1+=(p[i]==0),sum2+=(q[i]==0);
        printf("%d\n",max(sum1,sum2));
    }
    return 0;
}