HDU - 2767 Proving Equivalences(tarjan缩点)

本文解析了一道关于证明矩阵等价性的线性代数题目,通过Tarjan算法求解最少添加边数使图成为强连通图,并提供详细的代码实现。

题目

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:

给一个有向图,问最少添加几条边使得有向图成为一个强连通图。

思路:

Tarjan入门经典题,用tarjan缩点,然后就变成一个DAG了。
我们要考虑的问题是让它变成强连通,让DAG变成强连通就是把尾和头连起来,也就是入度和出度为0的点
统计DAG入度和出度,然后计算头尾,最大的那个就是所求

代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int maxn = 56789;
int T,n,m,tot,time,scc;
int cnt_in,cnt_out;
int head[maxn],dfn[maxn],low[maxn],belong[maxn];
int in[maxn],out[maxn];
stack<int>st;

struct DEGE
{
    int v;
    int next;
}e[maxn];

void init()
{
    cnt_in=cnt_out=time=scc=tot=0;
    memset(head,-1,sizeof(head));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
    memset(belong,0,sizeof(belong));
    while(!st.empty())
        st.pop();
}

void add_edge(int u,int v)
{
    tot++;
    e[tot].v=v;
    e[tot].next=head[u];
    head[u]=tot;
}

void dfs(int u)
{
    dfn[u]=low[u]=++time;
    st.push(u);
    for(int i=head[u];i!=-1;i=e[i].next)
    {
        int v=e[i].v;
        if(!dfn[v])
        {
            dfs(v);
            low[u]=min(low[u],low[v]);
        }
        else if(!belong[v])
        {
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(low[u]==dfn[u])
    {
        scc++;
        while(st.top()!=u)
        {
            int x=st.top();
            st.pop();
            belong[x]=scc;
        }
        belong[u]=scc;
        st.pop();
    }
}

void Get_point()
{
    for(int i=1;i<=n;i++)
    {
        for(int j=head[i];j!=-1;j=e[j].next)
        {
            int v=e[j].v;
            if(belong[i]!=belong[v])
            {
                out[belong[i]]++;
                in[belong[v]]++;
            }
        }
    }
}

void Find_scc()
{
    for(int i=1;i<=n;i++)
    {
        if(!dfn[i])
        {
            dfs(i);
        }
    }
}

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        init();
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            add_edge(u,v);
        }
        Find_scc();
        Get_point();
        for(int i=1;i<=scc;i++)
        {
            if(in[i]==0)
                cnt_in++;
            if(out[i]==0)
                cnt_out++;
        }
        if(scc==1)
            printf("0\n");
        else
            printf("%d\n",max(cnt_in,cnt_out));
    }
    return 0;
}
内容概要:本文系统介绍了算术优化算法(AOA)的基本原理、核心思想及Python实现方法,并通过图像分割的实际案例展示了其应用价值。AOA是一种基于种群的元启发式算法,其核心思想来源于四则运算,利用乘除运算进行全局勘探,加减运算进行局部开发,通过数学优化器加速函数(MOA)和数学优化概率(MOP)动态控制搜索过程,在全局探索与局部开发之间实现平衡。文章详细解析了算法的初始化、勘探与开发阶段的更新策略,并提供了完整的Python代码实现,结合Rastrigin函数进行测试验证。进一步地,以Flask框架搭建前后端分离系统,将AOA应用于图像分割任务,展示了其在实际工程中的可行性与高效性。最后,通过收敛速度、寻优精度等指标评估算法性能,并提出自适应参数调整、模型优化和并行计算等改进策略。; 适合人群:具备一定Python编程基础和优化算法基础知识的高校学生、科研人员及工程技术人员,尤其适合从事人工智能、图像处理、智能优化等领域的从业者;; 使用场景及目标:①理解元启发式算法的设计思想与实现机制;②掌握AOA在函数优化、图像分割等实际问题中的建模与求解方法;③学习如何将优化算法集成到Web系统中实现工程化应用;④为算法性能评估与改进提供实践参考; 阅读建议:建议读者结合代码逐行调试,深入理解算法流程中MOA与MOP的作用机制,尝试在不同测试函数上运行算法以观察性能差异,并可进一步扩展图像分割模块,引入更复杂的预处理或后处理技术以提升分割效果。
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