主题链接: http://acm.hdu.edu.cn/showproblem.php? pid=3836 Equivalent Sets Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others) Total Submission(s): 2890    Accepted Submission(s): 1006 Problem Description To prove two se…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3836 Equivalent Sets Description To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.Y…

Equivalent Sets Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others) Problem Description To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally…

Equivalent Sets Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others) Total Submission(s): 2798    Accepted Submission(s): 962 Problem Description To prove two sets A and B are equivalent, we can first prove A is a su…

Equivalent Sets Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others) Total Submission(s): 4782    Accepted Submission(s): 1721 Problem Description To prove two sets A and B are equivalent, we can first prove A is a s…

Problem Description To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent. You are to prove N sets are equivalent, using the method abo…

http://acm.hdu.edu.cn/showproblem.php?pid=3836 判断至少需要加几条边才能使图变成强连通 把图缩点之后统计入度为0的点和出度为0的点,然后两者中的最大值就是需要连的边, 例如,假设入度为0的点多,那么每次把出度为0的点连一条边指向入度为0的点,就构成了一个环, 所以构成了一个强连通分量,同理可得出度为0点多的情况. 这代码用g++ re了,c++才能ac. #include <iostream> #include <cstdio> #in…

题目大意:给出N个点,M条边.要求你加入最少的边,使得这个图变成强连通分量 解题思路:先找出全部的强连通分量和桥,将强连通分量缩点.桥作为连线,就形成了DAG了 这题被坑了.用了G++交的,结果一直RE,用C++一发就过了... #include <cstdio> #include <cstring> #define N 20010 #define M 100010 #define min(a,b) ((a) > (b)? (b): (a)) #define max(a,b)…

Equivalent Sets Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others)Total Submission(s): 3568    Accepted Submission(s): 1235 Problem Description To prove two sets A and B are equivalent, we can first prove A is a su…

http://acm.hdu.edu.cn/showproblem.php?pid=3836 Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others)Total Submission(s): 4802    Accepted Submission(s): 1725 Problem Description To prove two sets A and B are equivalen…

Problem Description To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent. You are to prove N sets are equivalent, using the method abo…

题目说给出一些子集,如果A是B的子集,B是A的子集,那么A和B就是相等的,然后给出n个集合m个关系,m个关系表示u是v的子集,问你最小再添加多少个关系可以让这n个集合都是相等的 如果这n个几个都是互相相等的,那么就等于是这n个集合看成点以后,构成的图是一个强连通图,那么就是说在加多少边让这个图变成强联通图. 先缩点然后做判断 1,如果原本的图本来就是强联通的,那么答案就是0 2.如果原本的图不是强联通的,那就去判断缩完以后的点的入度和出度,取入度为0的和出度为0的点数的最大值,就是最小需要加的边…

题意: 给一些集合 要求证明所有集合是相同的 证明方法是,如果$A∈B$,$B∈A$那么$A=B$成立 每一次证明可以得出一个$X∈Y$ 现在已经证明一些$A∈B$成立 求,最少再证明多少次,就可以完成要求 分析 其实就等价于给一个有向图,问你再加入多少个边可以使得图变为强连通图 给一个图论经典结论: "对于一个有向无环图(DAG),若想让它成为强连通图,至少需要添加$max(a,b)$条边 $a$为入度为0的点的数量,$b$为出度为0的点的数量" 而对于一个有向图,其每个强连通分量都…

题意: 给你一张有向图,问你最少加多少条边这张图强连通 思路: 缩点之后,如果不为1个点,答案为出度为0与入度为0的点的数量的最大值 代码: #include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> //#include<stack> #include<queue…

// 给你一个有向图,问你最少加几条边能使得该图强连通 #include <iostream> #include <cstdio> #include <cstring> #include <string> #include <utility> #include <algorithm> #include <vector> #include <queue> #include <stack> using…

#include<stdio.h>//求出其所有的强连通分量缩点,选出出度和入度最大的那个就是要求的边 #include<string.h> #include<stdlib.h> #define N 51000 struct node { int v,next; }bian[N]; int head[N],yong,n,indegree[N],outdegree[N],visit[N],suo[N],dfn[N],f,low[N],index,stac[N],top;…

刷题之前来几套LCA的末班 对于题目 HDU 2586 How far away 2份在线模板第一份倍增,倍增还是比较好理解的 #include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <…

转自:http://blog.csdn.net/shahdza/article/details/7779356 [HDU][强连通]:1269 迷宫城堡 判断是否是一个强连通★2767Proving Equivalences 至少加几条边让整个图变成强连通★3836 Equivalent Sets 至少加几条边让整个图变成强连通★1827 Summer Holiday 传递的最小费用★★3072 Intelligence System 传递的最小费用★★3861The King’s Proble…

转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并查集======================================[HDU]1213   How Many Tables   基础并查集★1272   小希的迷宫   基础并查集★1325&&poj1308  Is It A Tree?   基础并查集★1856   More i…

=============================以下是最小生成树+并查集====================================== [HDU] 1213 How Many Tables 基础并查集★ 1272 小希的迷宫 基础并查集★ 1325&&poj1308 Is It A Tree? 基础并查集★ 1856 More is better 基础并查集★ 1102 Constructing Roads 基础最小生成树★ 1232 畅通工程 基础并查集★ 123…

=============================以下是最小生成树+并查集====================================== [HDU] 1213 How Many Tables 基础并查集★ 1272 小希的迷宫 基础并查集★ 1325&&poj1308 Is It A Tree? 基础并查集★ 1856 More is better 基础并查集★ 1102 Constructing Roads 基础最小生成树★ 1232 畅通工程 基础并查集★ 123…

强连通 迷宫城堡 Proving Equivalences Equivalent Sets Summer Holiday Intelligence System The King's Problem Hawk-and-Chicken Cactus 仙人掌图 [双连通]: 2242 考研路茫茫--空调教室 双联通缩点+树形DP 2460 Network 边双连通 3849 By Recognizing These Guys, We Find Social Networks Useful 双连通求桥…

[图论01]最短路 Start Time : 2018-01-02 12:45:00    End Time : 2018-01-23 12:45:00 Contest Status : Running Current System Time : 2018-01-12 14:39:34 Solved Problem ID Title Ratio(Accepted / Submitted)   1001 最短路 51.85%(70/135)   1002 King 46.67%(35/75)  …

Equivalent Sets Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others) Total Submission(s): 2526    Accepted Submission(s): 857 Problem Description To prove two sets A and B are equivalent, we can first prove A is a su…

A.A + B problem(待填坑) B.Cat VS Dog(二分图匹配) 喜欢cat和喜欢dog的人构成了二分图,如果两个人有冲突则连一条边,则问题转化为二分图最大点独立集问题.ans=n-最大匹配. # include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue&g…

=============================以下是最小生成树+并查集====================================== [HDU] How Many Tables 基础并查集★ 小希的迷宫 基础并查集★ &&poj1308 Is It A Tree? 基础并查集★ More is better 基础并查集★ Constructing Roads 基础最小生成树★ 畅通工程 基础并查集★ 还是畅通工程 基础最小生成树★ 畅通工程 基础最小生成树★ 畅通…

这几天一直在做强连通,现在总结一小下 1.定义 在一个有向图中,如果任意的两个点都是相互可达的,就说这个图是强连通的,有向图的极大强连通子图,称为强连通分量 2.求法 学的是白书上的tarjan算法 用到了DFS的时间戳 假设一个强连通分量C,其中的第一个点是 P,那么DFS下去,就一定能够找到一个K点,返回P点,这条DFS路径上的点就处于这个强连通分量C中 假如现在发现节点v,同时发现节点v最远只能够到达节点u,那么节点u就是这个强连通分量最先被发现的节点 这样就转化成了求一个点u最远能够到达…

http://www.cnblogs.com/wenruo/p/4989425.html 强连通 强连通是指一个有向图中任意两点v1.v2间存在v1到v2的路径及v2到v1的路径. dfs遍历一个图,会生成一颗树.每个节点按最先遍历的时间给定一个编号,用一个数组dfn表示,又叫时间戳. 然后有几个概念. 画图举例: 假设一个边是u-->v 树边:dfs遍历后生成树的边叫做树边.dfn[u] = -1 如图中<1,2> <2,3> <3,4> <2,5>…

Tree Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others) Total Submission(s): 1643    Accepted Submission(s): 461 Problem Description   Zero and One are good friends who always have fun with each other. This time, t…

Caocao's Bridges Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3992    Accepted Submission(s): 1250 Problem Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi.…