UVA 12167 加最少的边让其变为连通图

ejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=&problem=3319&mosmsg=Submission+received+with+ID+13628687

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:
A is invertible.
Ax = b has exactly one solution for every n × 1 matrix b.
Ax = b is consistent for every n × 1 matrix b.
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

The 2008 ACM Northwestern European Programming Contest
图的连通性。在一个有向图上增加一些边，使其变为强连通图。先求强连通分量，将所有强连通分量缩为一个点，然后统计缩点后的图中入度为零和出度为零的点的个数，去最大即为所求。
#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
const int N=100005;
int head
,ip,in
,out
,stack
,ins
,index,dfn
,low
,ee
[2],belong
;
int cnt_tar,top,m,n;
struct note
{
int to;
int next;
}edge
;
void add(int u,int v)
{
edge[ip].to=v;
edge[ip].next=head[u];
head[u]=ip++;
}
void tarjan(int u)
{
int j,i,v;
dfn[u]=low[u]=++index;
stack[++top]=u;
ins[u]=1;
for(i=head[u]; i!=-1; i=edge[i].next)
{
v=edge[i].to;
if(!dfn[v])
{
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(ins[v])
low[u]=min(low[u],dfn[v]);
}
if(dfn[u]==low[u])
{
cnt_tar++;
do
{
j=stack[top--];
ins[j]=0;
belong[j]=cnt_tar;
}
while(j!=u);
}
}
void solve()
{
int i;
top=0,index=0,cnt_tar=0;
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
for(i=1; i<=n; i++)
{
if(!dfn[i])
{
tarjan(i);
}
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
memset(head,-1,sizeof(head));
memset(in,0,sizeof(in));
memset(out,0,sizeof(out));
scanf("%d%d",&n,&m);
for(int i=1; i<=m; i++)
{
scanf("%d%d",&ee[i][0],&ee[i][1]);
add(ee[i][0],ee[i][1]);
}
solve();
for(int i=1; i<=m; i++)
{
int xx=belong[ee[i][0]],yy=belong[ee[i][1]];
if(xx!=yy)
{
in[yy]++;
out[xx]++;
}
}
int innum=0,outnum=0;
for(int i=1; i<=cnt_tar; i++)
{
if(in[i]==0)
innum++;
if(out[i]==0)
outnum++;
}
printf("%d\n",max(innum,outnum));
}
return 0;
}