HDU2767 Proving Equivalences 解题报告【tarjan缩点】

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
解题报告
这道题的题意是问我们一幅图里边，至少添加多少条边可以使整个图变成一个强连通块。
显然我们第一步应该用tarjan缩点。
有两个特判，若是边数为0，那么就输出点数，若是缩点后只有一个点，那么就输出0。
除开特判，我们要统计出缩点后每个点的出度入度，然后分别统计入度或出度为0的点的个数，输出两者中的最大值。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
using namespace std;
const int N=20000,M=50000;
struct edge
{
    int v,next;
}ed[M+5];
int T,n,m,tim,cnt,top;
int head[N+5],num;
int in[N+5],out[N+5];
int dfn[N+5],low[N+5],bl[N+5],sta[N+5],flag[N+5];
void build(int u,int v)
{
    ed[++num].v=v;
    ed[num].next=head[u];
    head[u]=num;
}
void tarjan(int u)
{
    dfn[u]=low[u]=++tim;
    flag[sta[++top]=u]=1;
    for(int i=head[u];i!=-1;i=ed[i].next)
    {
        int v=ed[i].v;
        if(!dfn[v])
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(flag[v])low[u]=min(low[u],dfn[v]);
    }
    if(dfn[u]==low[u])
    {
        cnt++;
        for(int now=0;now!=u;)
        {
            flag[now=sta[top--]]=0;
            bl[now]=cnt;
        }
    }
}
int main()
{
    for(scanf("%d",&T);T;T--)
    {
        scanf("%d%d",&n,&m);
        if(m==0){printf("%d\n",n);continue;}
        tim=0,cnt=0,top=0,num=0;
        memset(head,-1,sizeof(head));
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        memset(bl,0,sizeof(bl));
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        memset(flag,0,sizeof(flag));
        while(m--)
        {
            int u,v,w;
            scanf("%d%d",&u,&v);
            build(u,v);
        }
        for(int i=1;i<=n;i++)if(!dfn[i])tarjan(i);
        if(cnt==1){printf("0\n");continue;}
        for(int u=1;u<=n;u++)
        for(int i=head[u];i!=-1;i=ed[i].next)
        {
            int v=ed[i].v;
            if(bl[u]!=bl[v])in[bl[v]]++,out[bl[u]]++;
        }
        int ans1=0,ans2=0;
        for(int i=1;i<=cnt;i++)
        {
            if(in[i]==0)ans1++;
            if(out[i]==0)ans2++;
        }
        printf("%d\n",max(ans1,ans2));
    }
    return 0;
}