给一个图,问至少加入�多少条有向边能够使图变成强连通的. 原图是有环的,缩点建图,在该DAG图上我们能够发现,要使该图变成强连通图必须连成环 而加入�最少的边连成环,就是把图上入度为0和出度为0的点连上,那么其它的点就都能够互相到达了 所以答案就是max(入度为0的点,出度为0的点) #include <iostream> #include <cstring> #include <string> #include <cstdio> #include <…

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…

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…

题意: 给一个有向图,问添加几条边可以使其强连通. 思路: tarjan算法求强连通分量,然后缩点求各个强连通分量的出入度,答案是max(入度为0的缩点个数,出度为0的缩点个数). #include <bits/stdc++.h> #define LL long long #define pii pair<int,int> using namespace std; +; const int INF=0x7f7f7f7f; vector<int> vect[N]; sta…

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&…

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5742    Accepted Submission(s): 1973 Problem Description Consider the following exercise, found in a generic linear algebra textbook. Let A be an…

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 &…

http://acm.hdu.edu.cn/showproblem.php?pid=2767 求至少添加多少条边才能变成强连通分量.统计入度为0的点和出度为0的点,取最大值即可. #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <algorithm> #include <str…

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…

原题链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2288 题意: 给你一个有向图,问你至少需要添加多少条边,使得整个图强连通. 题解: 就..直接缩点,令缩点后入度为0的点有a个,出度为0的点有b个,答案就是max(a,b) 代码: #include<iostream> #include<cstri…

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 题目链接(点击) 参考博客(点击) Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9497    Accepted Submission(s): 3349Problem Description Consider the following exercise, found in a generi…

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,…

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…

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 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…

Proving Equivalences Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6006    Accepted Submission(s): 2051 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): 4263    Accepted Submission(s): 1510 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): 4384    Accepted Submission(s): 1556 Problem Description Consider the following exercise, found in a generic linear algebra t…

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…

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…

[题目链接]hdu-2767 [题目描述] 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 ma…

http://acm.hdu.edu.cn/showproblem.php?pid=2767 题意:给出n个点m条边,问在m条边的基础上,最小再添加多少条边可以让图变成强连通.思路:强连通分量缩点后找入度为0和出度为0的点,因为在强连通图里面没有一个点的入度和出度都为0,所以取出度为0的点和入度为0的点中的最大值就是答案.(要特判强连通分量数为1的情况) #include <cstdio> #include <algorithm> #include <iostream>…

题意:有n个命题,已知其中的m个推导,要证明n个命题全部等价(等价具有传递性),最少还需要做出几次推导. 思路:由已知的推导可以建一张无向图,则问题变成了最少需要增加几条边能使图变成强连通图.找出所有的强连通分量,将每一个连通分量视作一个大节点,则整张图变成了一张DAG.设出度为0的大节点个数为b,入度为0的大节点个数为a,则答案就是max(a,b). #include<iostream> #include<string> #include<algorithm> #in…

就是统计入度为0 的点 和 出度为0 的点  输出 大的那一个,, 若图中只有一个强连通分量 则输出0即可 和https://www.cnblogs.com/WTSRUVF/p/9301096.html  这题差不多  poj1236 #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set…

题目大意:有n个命题,已知其中的m个推导,要证明n个命题全部等价(等价具有传递性),最少还需要做出几次推导. 题目分析:由已知的推导可以建一张无向图,则问题变成了最少需要增加几条边能使图变成强连通图.找出所有的强连通分量,将每一个连通分量视作一个大节点,则整张图变成了一张DAG.设出度为0的大节点个数为a,入度为0的大节点个数为b,则答案就是max(a,b).为什么是这样呢?因为要使等价证明前进下去,每个大节点的出度和入度都必须不能是0. 代码如下: # include<iostream> #…

题意:给出一个有向图,问最少添加几条有向边使得原图强连通. 解法:求出SCC后缩点,统计一下出度为0的点和入度为0的点,二者取最大值就是答案. 还有个特殊情况就是本身就是强连通的话,答案就是0. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <stack> using na…

题意:给定一个图,问至少加入多少条边能够使这个图强连通. 思路:首先求出这个图的强连通分量.然后把每个强连通分量缩成一个点.那么这个图变成了一个DAG,求出全部点的入度和出度,由于强连通图中每个节点的入度和出度至少为1.那么我们求出入度为零的节点数量和出度为零的节点数量.答案取最大值,由于在一个DAG中加入这么多边一定能够使这个图强连通.注意当这个图本身强连通时要特判一下,答案为零. #include<cstdio> #include<cstring> #include<cm…

#include<stdio.h> #include<string.h> #define N 21000 struct node { int u,v,next; }bian[N*3]; int head[N],dfn[N],low[N],index,vis[N],stac[N],top,yong,cnt,suo[N],indegree[N],outdegree[N]; void init() { yong=0;index=0;top=0;cnt=0; memset(vis,0,si…