hdu2767——Proving Equivalences

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最新推荐文章于 2024-08-22 20:10:34 发布

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本文探讨了一种算法问题，即通过已知的一些命题间的蕴含关系，确定最少还需证明多少额外的蕴含关系才能使所有命题互相等价。利用Tarjan算法进行强连通分量计算，并在此基础上进一步分析每个强连通分量的进出关系来解决问题。

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Proving Equivalences

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3226 Accepted Submission(s): 1216

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

Sample Output

4
2

Source

NWERC 2008

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水题，就是问你最少加几条边使得图变成强连通图，tarjan+缩点，计算入度和出度为0的点的个数，然后找大的那个，注意如果已经是强连通图了，答案就是0了，特判一下，我这里没注意WA了3发 Orz

#include<map>
#include<set>
#include<list>
#include<stack>
#include<queue>
#include<vector>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

const int N  = 20010;
const int M  = 50010;
int DFN[N];
int low[N];
int block[N];
int Stack[N];
int out[N];
int in[N];
bool instack[N];
int head[N];
int tot, sccnum, index, top, n, m;

struct node
{
    int next;
    int to;
}edge[M];

void addedge(int from, int to)
{
    edge[tot].to = to;
    edge[tot].next = head[from];
    head[from] = tot++;
}

void tarjan(int u)
{
    DFN[u] = low[u] = ++index;
    Stack[top++] = u;
    instack[u] = 1;
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if (DFN[v] == 0)
        {
            tarjan(v);
            if (low[u] > low[v])
            {
                low[u] = low[v];
            }
        }
        else if (instack[v])
        {
            if (low[u] > DFN[v])
            {
                low[u] = DFN[v];
            }
        }
    }
    if (DFN[u] == low[u])
    {
        sccnum++;
        do
        {
            top--;
            block[Stack[top]] = sccnum;
            instack[Stack[top]] = 0;
        }while ( top >=0 && Stack[top] != u);
    }
}

void solve()
{
    memset( instack, 0, sizeof(instack) );
    memset( DFN, 0, sizeof(DFN) );
    memset( low, 0, sizeof(low) );
    memset( in, 0, sizeof(in) );
    memset( out, 0, sizeof(out) );
    sccnum = index = top = 0;
    for (int i = 1; i <= n; i++)
    {
        if (DFN[i] == 0)
        {
            tarjan(i);
        }
    }
    if(sccnum == 1)
    {
        printf("0\n");
        return ;
    }
    for (int u = 1; u <= n; u++)
    {
        for (int j = head[u]; j != -1; j = edge[j].next)
        {
            int v = edge[j].to;
            if (block[v] != block[u])
            {
                out[block[u]]++;
                in[block[v]]++;
            }
        }
    }
    int a = 0, b = 0;
    for (int u = 1; u <= sccnum; u++)
    {
        if (in[u] == 0)
        {
            b++;
        }
        if (out[u] == 0)
        {
            a++;
        }
    }
    printf("%d\n", max(a, b));
}

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        memset( head, -1, sizeof(head) );
        tot = 0;
        scanf("%d%d", &n, &m);
        int u, v;
        for (int i = 0; i < m; i++)
        {
            scanf("%d%d", &u, &v);
            addedge(u, v);
        }
        solve();
    }
    return 0;
}