hdoj 2767 Proving Equivalences【强连通&&tarjan】

Proving Equivalences

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

Total Submission(s): 4407 Accepted Submission(s): 1560

[/b]

[align=left]Problem Description[/align]
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?

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2
4 0
3 2
1 2
1 3

[align=left]Sample Output[/align]

4
2

分析：
强连通果题，先压缩路径，把线路转化为点，判断点的出度和入度。
代码：

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<stack>
#include<vector>
using namespace std;
#define maxn 20010
vector<int>g[maxn];
stack<int>s;
int dfn[maxn],low[maxn],sccno[maxn],tclock,scccnt;
int indeg[maxn],outdeg[maxn];
int max(int a,int b)
{
return a>b?a:b;
}

void tarjan(int u)
{
dfn[u]=low[u]=++tclock;
s.push(u);
for(int i=0;i<g[u].size();i++)
{
int v=g[u][i];
if(!dfn[v])
{
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(!sccno[v])
{
low[u]=min(low[u],dfn[v]);
}
}
if(low[u]==dfn[u]){
scccnt+=1;
int v=-1;
while(v!=u)
{
v=s.top();
s.pop();
sccno[v]=scccnt;
}
}
}

void findscc(int n)
{
tclock=scccnt=0;
memset(dfn,0,sizeof(int)*(n+1));
memset(sccno,0,sizeof(int)*(n+1));
memset(low,0,sizeof(int)*(n+1));
for(int i=1;i<=n;i++)
if(!dfn[i])
tarjan(i);
}

int workout(int n)
{
if(scccnt==1)
return 0;
memset(indeg,0,sizeof(int)*(n+1));
memset(outdeg,0,sizeof(int)*(n+1));
for(int u=1;u<=n;u++)
{
for(int i=0;i<g[u].size();i++)
{
int v=g[u][i];
if(sccno[u]!=sccno[v])
{
indeg[sccno[v]]+=1;
outdeg[sccno[u]]+=1;
}
}
}
int c1=0,c2=0;
for(int i=1;i<=scccnt;i++){
if(indeg[i]==0)
c1++;
if(outdeg[i]==0)
c2++;}
return max(c1,c2);
}

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
g[i].clear();
for(int i=0;i<m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
g[a].push_back(b);
}
findscc(n);
printf("%d\n",workout(n));
}
return 0;
}