Proving Equivalences

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

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

题意：有n个命题，现已给出m次推导即由a可以推出b（根据该条件b不能推出a），问最少还需要多少次推导才可以证明n个命题之间是等价的。（等价命题：a <=> b 即 a可以推出b 且 b可以推出a。）

把每个命题虚拟成一个节点，a推导b当作一条a到b的有向边，问题就变成：在有向图中最少增加多少条边才可以使新图强连通。

思路：

将【每个强连通分支】缩成一个点，记录每个【缩点】的出度， 入度，并构造新图，然后求出所有入度为0的点数numin，求出所有出度为0的点数numuot，取最大值即可

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
#define maxn 20000 + 1000
#define maxm 50000 + 5000
using namespace std;
int n, m;
vector<int>edge[maxm];
int low[maxn], dfn[maxn];
int dfs_clock;
int Stack[maxn];
bool Instack[maxn];
int top;
int Belong[maxn] , scc_clock;
int out[maxn], in[maxn];

void init(){
    for(int i = 1; i <= n; ++i)
        edge[i].clear();
}

void getmap(){
    scanf("%d%d", &n, &m);
    while(m--){
        int a, b;
        scanf("%d%d", &a, &b);
        edge[a].push_back(b);
    }
}

void tarjan(int u, int per){
    int v;
    low[u] = dfn[u] = ++dfs_clock;
    Stack[top++] = u;
    Instack[u] = true;
    for(int i = 0; i < edge[u].size(); i++){
        v = edge[u][i];
        if(!dfn[v]){
            tarjan(v, u);
            low[u] = min(low[v], low[u]);
        }
        else if(Instack[v]){
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(dfn[u] == low[u]){
        scc_clock++;
        do{
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = scc_clock;
        }while(u != v);
    }
}

void find(){
    memset(low, 0, sizeof(low));
    memset(dfn, 0, sizeof(dfn));
    memset(Instack, false, sizeof(Instack));
    memset(Belong, 0, sizeof(Belong));
    dfs_clock = scc_clock = top = 0;
    for(int i = 1; i <= n; ++i){
        if(!dfn[i])
            tarjan(i, i);
    }
}

void suodian(){
    for(int i = 1; i <= scc_clock; ++i){
        in[i] = 0;
        out[i] = 0;
    }
    for(int i = 1; i <= n; ++i){
        for(int j = 0; j < edge[i].size(); ++j){
            int u = Belong[i];
            int v = Belong[edge[i][j]];
            if(u != v){
                out[u]++, in[v]++;
            }
        }
    }
}

void solve(){
    if(scc_clock == 1){
        printf("0\n");
        return ;
    }
    int numin = 0, numout = 0;
    for(int i = 1; i <= scc_clock; ++i){
        if(in[i] == 0) numin++;
        if(out[i] == 0) numout++;
    }
    //printf("%d\n",scc_clock);
    printf("%d\n", max(numin, numout));
}

int main (){
    int T;
    scanf("%d", &T);
	while(T--){
        init();
        getmap();
        find();
        suodian();
        solve();
	}
	return 0;
}

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