Proving Equivalences Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4384    Accepted Submission(s): 1556 Problem Description Consider the following exercise, found in a generic linear algebra t…

Proving Equivalences Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10665    Accepted Submission(s): 3606 Problem Description Consider the following exercise, found in a generic linear algebra…

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…

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: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3743    Accepted Submission(s): 1374 Problem Description Consider the following exercise, found in a generic linear algebra…

点击打开链接 有向图强联通,Kosaraju算法 缩点后分别入度和出度为0的点的个数 answer = max(a, b); scc_cnt = 1; answer = 0 #include<cstdio> #include<algorithm> #include<vector> #include<cstring> #include<stack> using namespace std; const int maxn = 20000 + 10;…

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…

UVA12167 Proving Equivalences 题意翻译 题目描述 在数学中,我们常常需要完成若干命题的等价性证明. 例如:有4个命题a,b,c,d,要证明他们是等价的,我们需要证明a<=>b,然后b<=>c,最后c<=>d.注意每次证明是双向的,因此一共完成了6次推导.另一种证明方法是:证明a->b,然后b->c,接着c->d,最后d->a,只须4次证明. 现在你任务是证明 n 个命题全部等价,且你的朋友已经为你作出了m次推导(已知…

题目地址:pid=2767">HDU 2767 题意:给一张有向图.求最少加几条边使这个图强连通. 思路:先求这张图的强连通分量.假设为1.则输出0(证明该图不须要加边已经是强连通的了).否则缩点. 遍历原图的全部边.假设2个点在不同的强连通分量里面,建边,构成一张新图.统计新图中点的入度和出度,取入度等于0和出度等于0的最大值(由于求强连通缩点后.整张图就变成了一个无回路的有向图.要使之强连通.仅仅须要将入度=0和出度=0的点加边就可以,要保证加边后没有入度和出度为0的点,所以取两者最大…

Problem   UVALive - 4287 - Proving Equivalences Time Limit: 3000 mSec Problem Description Input Output Sample Input 2 4 0 3 2 1 2 1 3 Sample Output 4 2 题解:题意就是给出一个有向图,问最少添加几条有向边能够使得整张图强连通,Tarjan缩点是比较容易想到的,之后怎么办,要用到一个结论:如果图中有a个入度为零的点,b个出度为零的点,那么max(a,…