Proving Equivalences

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

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

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=20000+10;
int n,m,size,top,index,ind,oud;
int head[MAX],dfn[MAX],low[MAX],stack[MAX];
int mark[MAX],flag[MAX];
//dfn表示点u出现的时间,low表示点u能到达所属环中最早出现的点(记录的是到达的时间) 

struct Edge{
	int v,next;
	Edge(){}
	Edge(int V,int NEXT):v(V),next(NEXT){}
}edge[50000+10];

void Init(int num){
	for(int i=0;i<=num;++i)head[i]=-1;
	size=top=index=ind=oud=0;
}

void InsertEdge(int u,int v){
	edge[size]=Edge(v,head[u]);
	head[u]=size++;
}

void tarjan(int u){
	if(mark[u])return;
	dfn[u]=low[u]=++index;
	stack[++top]=u;
	mark[u]=1;
	for(int i=head[u];i != -1;i=edge[i].next){
		int v=edge[i].v;
		tarjan(v);
		if(mark[v] == 1)low[u]=min(low[u],low[v]);//必须点v在栈里面才行
	}
	if(dfn[u] == low[u]){
		++ind,++oud;//计算缩点后点的个数,方便计算入度和出度
		while(stack[top] != u){
			mark[stack[top]]=-1;
			low[stack[top--]]=low[u];
		}
		mark[u]=-1;
		--top;
	}
}

int main(){
	int t,u,v;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&m);
		Init(n);
		for(int i=0;i<m;++i){
			scanf("%d%d",&u,&v);
			InsertEdge(u,v);
		}
		memset(mark,0,sizeof mark);
		for(int i=1;i<=n;++i){
			if(mark[i])continue;
			tarjan(i);//tarjan用来缩点
		}
		if(ind == 1){cout<<0<<endl;continue;}
		for(int i=0;i<=n;++i)mark[i]=flag[i]=0;
		for(int i=1;i<=n;++i){
			for(int j=head[i];j != -1;j=edge[j].next){
				v=edge[j].v;
				if(low[i] == low[v])continue;
				if(mark[low[i]] == 0)--oud;//mark标记点u是否有出度
				if(flag[low[v]] == 0)--ind;//flag标记点u是否有入度
				mark[low[i]]=1,flag[low[v]]=1;
			}
		}
		printf("%d\n",max(oud,ind));
	}
	return 0;
}

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