【HDU 2767】Proving Equivalences （Tarjan 缩点）

Proving Equivalences

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5184 Accepted Submission(s): 1811

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

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[题意][给定一个有向图和一些边，计算还需加入几条边使之成为一个强连通分量]

【题解】【Tarjan算法，缩点，多组数据，别忘清零。】

【几个特判：①没有给边的情况：直接输出n；②缩点后就剩一个点的情况：输出零】

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[500010],nxt[500010],p[200010],tot;
int dft[200010],dis[200010],f[200010],cnt,root;
int in[200010],out[200010];
int n,m,N,d[200010],top;
bool vis[200010];

inline void add(int x,int y)
{
	tot++; a[tot]=y; nxt[tot]=p[x]; p[x]=tot;
}
inline void clear()
{
	memset(nxt,-1,sizeof(nxt)); memset(p,-1,sizeof(p));
	memset(dft,0,sizeof(dft));  memset(dis,0,sizeof(dis));
	memset(vis,0,sizeof(vis));	memset(d,0,sizeof(d));
	memset(out,0,sizeof(out));  memset(in,0,sizeof(in));
	memset(f,0,sizeof(f));
	tot=cnt=root=top=0;
}

void tarjan(int x)
{
	dft[x]=dis[x]=++cnt;
	d[++top]=x; vis[x]=1;
	int u=p[x];
	while(u>=0)
	 {
	 	if(!dft[a[u]])
	 	  tarjan(a[u]),dis[x]=min(dis[x],dis[a[u]]);
	 	 else 
	 	   if(vis[a[u]]) dis[x]=min(dis[x],dft[a[u]]);
	 	u=nxt[u];
	 }
	if(dis[x]==dft[x])
	 {
	 	root++;
	 	do{
	 		u=d[top--];
	 		vis[u]=0;
	 		f[u]=root;
	 	}while(u!=x);
	 }
	return;
}
inline void find()
{
	for(int i=1;i<=n;++i)
	 {
	 	int u=p[i];
	 	while(u>=0)
	 	 {
	 	 	if(f[a[u]]!=f[i])
	 	 	  out[f[i]]++,in[f[a[u]]]++;
	 	 	u=nxt[u];
	 	 }
	 }
	return;
}

int main()
{
	int i,j;
	scanf("%d",&N);
	for(i=1;i<=N;++i)
	 {
	    clear();
	 	scanf("%d%d",&n,&m);
	 	if(!m) {printf("%d\n",n); continue;}
	 	for(j=1;j<=m;++j)
	 	 {
	 	 	int x,y;
	 	 	scanf("%d%d",&x,&y);
	 	 	add(x,y);
	 	 }
	 	for(j=1;j<=n;++j)
	 	 if(!dft[j]) tarjan(j);
	 	if(root==1) {printf("0\n"); continue;}
		find(); 
		int inr=0,outr=0;
		for(j=1;j<=root;++j)
		 {
		 	if(!in[j]) inr++;
		 	if(!out[j]) outr++;
		 }
		printf("%d\n",max(inr,outr));
	 }
	return 0;
}