hdu2767（强连通分量+缩点）

Proving Equivalences

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7367 Accepted Submission(s): 2547

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
对整个图求一次强连通分量，如果强连通分量为1则直接输出0，否则进行缩点（啥叫缩点：我们求强连通分量时，给每个顶点做一个标记，标记该顶点属于哪个强联通分量，然后属于同一个强连通分量的点就可以看作同一个点了。这就是所谓的“缩点”）对整个图缩点后这个图就变成了有向无环图，假设这个有向无环图入度为零的点有a个，出度为零的点有b个，这结果为max(a,b)（这个结论可以画个图推一推）

#include <bits/stdc++.h>
using namespace std;
const int N=20010;
vector <vector<int> > vec(N);
int low[N],dfn[N],Stack[N],belong[N];
bool InStack[N];
int in[N],out[N];
int Index,top,ans;
void Tarjan(int u){
    low[u]=dfn[u]=(++Index);
    Stack[top++]=u;
    InStack[u]=true;
    for(int i=0;i<vec[u].size();i++){
        int v=vec[u][i];
        if(!dfn[v]){
            Tarjan(v);
            if(low[u]>low[v]){
                low[u]=low[v];
            }
        } else if(InStack[v]&&low[u]>dfn[v]){
            low[u]=dfn[v];
        }
    }
    if(low[u]==dfn[u]){
        int v;
        ans++;
        do{
            v=Stack[--top];
            belong[v]=ans;
            InStack[v]=false;
        } while(v!=u);
    }
}
void init(int n){
    for(int i=1;i<=n;i++){
        vec[i].clear();
    }
    memset(InStack,false,sizeof(InStack));
    memset(belong,0,sizeof(belong));
    memset(dfn,0,sizeof(dfn));
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
    Index=top=ans=0;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n,m;
        scanf("%d %d",&n,&m);
        init(n);
        for(int i=0;i<m;i++){
            int u,v;
            scanf("%d %d",&u,&v);
            vec[u].push_back(v);
        }
        for(int i=1;i<=n;i++){
            if(!dfn[i]){
                Tarjan(i);
            }
        }
        if(ans==1){
            puts("0");
        } else {
            for(int i=1;i<=n;i++){
                for(int j=0;j<vec[i].size();j++){
                    int v=vec[i][j];
                    if(belong[i]!=belong[v]){
                        in[belong[v]]++;
                        out[belong[i]]++;
                    }
                }
            }
            int a=0,b=0;
            for(int i=1;i<=ans;i++){
                if(!in[i]) a++;
                if(!out[i]) b++;
            }
            printf("%d\n",max(a,b));
        }
    }
    return 0;
}