给出n个命题,m个推导,问最少添加多少条推导,能够使全部命题都能等价(两两都能互推) 既给出有向图,最少加多少边,使得原图变成强连通. 首先强连通缩点,对于新图,每一个点都至少要有一条出去的边和一条进来的边(这样才干保证它能到随意点和随意点都能到它) 所以求出新图中入度为0的个数,和出度为0的个数,加入的边就是从出度为0的指向入度为0的.这样还会有一点剩余,剩余的就乱连即可了. 所以仅仅要求出2者的最大值就OK. #include <iostream> #include<cstring&…

pid=2767">http://acm.hdu.edu.cn/showproblem.php?pid=2767 Proving Equivalences Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2926    Accepted Submission(s): 1100 Problem Description Conside…

Proving Equivalences Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 3   Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Consider the follo…

Proving Equivalences 题意:输入一个有向图(强连通图就是定义在有向图上的),有n(1 ≤ n ≤ 20000)个节点和m(0 ≤ m ≤ 50000)条有向边:问添加几条边可使图变成强连通图: 强连通分量:对于分量中的任意两个节点,都存在一条有向的路径(顺序不同,表示的路径不同):说白了,就是任意两点都能形成一个环(但不是说只有一个环) 思路:使用Tarjan算法 (讲解得很好)即可容易地在得到在一个强连通分量中设置每个点所属的强连通分量的标号:即得到所谓的缩点:之后就是合并…

Proving Equivalences Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9296    Accepted Submission(s): 3281 Problem Description Consider the following exercise, found in a generic linear algebra t…

题意: 一个有向图,问最少加几条边,能让它强连通 方法: 1:tarjan 缩点 2:采用如下构造法: 缩点后的图找到所有头结点和尾结点,那么,可以这么构造:把所有的尾结点连一条边到头结点,就必然可以是强连通了.如果说有几个结点不连通,那么让他们的尾结点相互只向对方的头结点就好了. 那么,最后的答案就是,头结点和尾结点中比较小的那个数量. 当然,如果缩点后只有一个点,那么就是0; 代码: #include <cstdio> #include <cstring> #include &…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2767 题意:问最少加多少边可以让所有点都相互连通. 题解:如果强连通分量就1个直接输出0,否者输出入度为0的缩点,出度为0的缩点和的最大值 #include <iostream> #include <cstring> #include <cstdio> using namespace std; const int N = 2e4 + 10; const int M = 5e…

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 c…

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4343    Accepted Submission(s): 1541 Problem Description Consider the following exercise, found in a generic linear algebra textbook. Let A be an…

Proving Equivalences Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4263    Accepted Submission(s): 1510 Problem Description Consider the following exercise, found in a generic linear algebra t…