B. Bear and Friendship Condition time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Bear Limak examines a social network. Its main functionality is that two members can become friends (then th…

B. Bear and Friendship Condition time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Bear Limak examines a social network. Its main functionality is that two members can become friends (then th…

B. Bear and Friendship Condition time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bear Limak examines a social network. Its main functionality is that two members can become friends (then th…

B. Bear and Friendship Condition time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bear Limak examines a social network. Its main functionality is that two members can become friends (then th…

B. Bear and Friendship Condition 题目连接: http://codeforces.com/contest/791/problem/B Description Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pic…

Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures). There are n members, numbered 1 through n. m pairs of members are friends. Of course, a…

[题目链接]:http://codeforces.com/contest/791/problem/B [题意] 给你m对朋友关系; 如果x-y是朋友,y-z是朋友 要求x-z也是朋友. 问你所给的图是否符合 [题解] 用并查集处理出每个朋友"团"的大小->就是连通块 然后这个连通块里面应该要有x*(x-1)/2条边; 按照这个规则,求出应该有的边数; 然后和所给的m对比; 相同则YES,否则NO [完整代码] #include <bits/stdc++.h> usin…

题目大意:给定一张无向图,要求如果 A 与 B 之间有边,B 与 C 之间有边,那么 A 与 C 之间也需要有边.问这张图是否满足要求. 题解:根据以上性质,即:A 与 B 有关系,B 与 C 有关系,那么 A 和 C 也要有关系,因此可以采用并查集加以维护,维护关系的同时顺便维护各个联通块的大小,若符合题目要求,则同一个联通块中的点必须均有关系.因此,最后计算一下每个联通块的应有关系数和最初所给的关系数比较,相等则符合,反之,不符合. 代码如下 #include <bits/stdc++.h>…

题意:如果1认识2,2认识3,必须要求有:1认识3.如果满足上述条件,输出YES,否则输出NO. 思路:显然如果是一个完全图就输出YES,否则就输出NO,如果是无向完全图则一定有我们可以用dfs来书边和点 n个节点的有向完全图边数为e=n*(n-1) 代码: #include <bits/stdc++.h> #define maxn 150000 #define ll long long using namespace std; vector <]; ]; int t; void dfs…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2833 A friend is like a flower, a rose to be exact, Or maybe like a brand new gate that never comes unlatched. A friend is like an owl, both beautiful and wise. Or perhaps a friend is like…