hdu2767之強聯通縮點

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Proving Equivalences Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2768    Accepted Submission(s): 1038


Problem DescriptionConsider 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? 
InputOn 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. 
OutputPer 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

24 03 21 21 3

 
Sample Output

42

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <queue>#include <algorithm>#include <map>#include <cmath>#include <iomanip>#define INF 99999999typedef 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表示點u出現的時間,low表示點u能到達所屬環中最早出現的點(記錄的是到達的時間) 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]);//必須點v在棧裡面才行 }if(dfn[u] == low[u]){++ind,++oud;//計算縮點後點的個數,方便計算入度和出度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用來縮點 }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標記點u是否有出度if(flag[low[v]] == 0)--ind;//flag標記點u是否有入度mark[low[i]]=1,flag[low[v]]=1; }}printf("%d\n",max(oud,ind));}return 0;}