hdu-2767

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

24 03 21 21 3

Sample Output

42

题意：抽象题意，去除那些废话后，剩下的意思就是问最少添加多少条边使得图变成强连通。（模板题！模板题！）

题解：先求强连通分量，再把强连通分量缩成DAG，然后求出入度为0的DAG个数的最大值。这个最大值就是答案。（结果我因为一个++tot和tot++搞了一个下午，最后找到症结内心是崩溃的，然而，还是有好处的，下次就不会再犯咯）

代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <stack>#include <cstring>#include <vector>#include <bitset>#include <string>#include <cmath>#include <set>#include <map>#include <queue>using namespace std;typedef long long ll;const int maxn = 2e4+7;vector<int> g[maxn];stack<int> s;int n,m;int low[maxn],dfn[maxn],vis[maxn],id[maxn];int ind[maxn],oud[maxn];int tot,cnt;void tarjan(int x){    low[x] = dfn[x] = ++tot;    s.push(x),vis[x] =1;    for (int i = 0;i < g[x].size();i++){        int v = g[x][i];        if(!dfn[v]){            tarjan(v);            low[x] = min(low[v],low[x]);        }        else if(vis[v]){            low[x] = min(low[x],dfn[v]);        }    }    if(low[x]==dfn[x]){        cnt++;        while (1){            int now = s.top();            s.pop();            vis[now] = 0;            id[now] = cnt;            if(now==x) break;        }    }}int main(){    int t;    scanf("%d",&t);    while(t--){        scanf("%d %d",&n,&m);        for (int i = 1;i <= n;i++) g[i].clear();        memset(dfn,0,sizeof(dfn));        memset(low,0,sizeof(low));        memset(vis,0,sizeof(vis));        memset(id,0,sizeof(id));        tot = 0,cnt = 0;        for (int i = 0;i < m;i++){            int a,b;            scanf("%d %d",&a,&b);            g[a].push_back(b);        }        for (int i = 1;i <= n;i++)        {            if(!dfn[i]){                tarjan(i);            }        }        if(cnt==1){            printf("%d\n",0);            continue;        }        memset(ind,0,sizeof(ind));        memset(oud,0,sizeof(oud));        for(int i = 1;i <= n;i++){            for (int j = 0;j < g[i].size();j++){                int u = id[i],v = id[g[i][j]];                if(u!=v){                    ind[v] = oud[u] = 1;                }            }        }        int ans1 = 0,ans2 = 0;        for (int i = 1;i <= cnt;i++){            if(ind[i]==0) ans1++;            if(oud[i]==0) ans2++;        }        printf("%d\n",max(ans1,ans2));    }    return 0;}