\(\color{#0066ff}{题目描述}\) 给出n个点,n-1条边,求再最多再添加多少边使得二分图的性质成立 \(\color{#0066ff}{输入格式}\) The first line of input contains an integer n - the number of nodes in the tree ( \(1<=n<=10^{5}\) ). The next n−1 lines contain integers u and v ( \(1<=u,v<=…

 Mahmoud and Ehab and the bipartiteness Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees. A tree is a connected acyclic graph. A bipartite graph is a graph, whose verti…

862B - Mahmoud and Ehab and the bipartiteness 思路:先染色,然后找一种颜色dfs遍历每一个点求答案. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) ; bool color[N]; vector<int>g[N]; ; int…

题意:给定一个n个点的树,该树同时也是一个二分图,问最多能添加多少条边,使添加后的图也是一个二分图. 分析: 1.通过二分图染色,将树中所有节点分成两个集合,大小分别为cnt1和cnt2. 2.两个集合间总共可以连cnt1*cnt2条边,给定的是一个树,因此已经连了n-1条边,所以最多能连cnt1*cnt2-(n-1)条边. 3.注意输出. #include<cstdio> #include<cstring> #include<cstdlib> #include<…

Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees. A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in su…

http://codeforces.com/problemset/problem/862/B 题意: 给出一个有n个点的二分图和n-1条边,问现在最多可以添加多少条边使得这个图中不存在自环,重边,并且此图还是一个二分图. 思路: 想得比较复杂了....其实既然已经给出了二分图并且有n-1条边,那么我们就一定可以用染色法对每一个点进行染色,从而将点划分为两个集合,然后从两个集合中分别任意选择一个点连边就行了. 一开始考虑二分图的基本属性是不存在奇数条边的环...并不需要这样,因为两个集合是分开的,…

[链接]h在这里写链接 [题意] 让你在一棵树上,加入尽可能多的边. 使得这棵树依然是一张二分图. [题解] 让每个节点的度数,都变成二分图的对方集合中的点的个数就好. [错的次数] 0 [反思] 在这了写反思 [代码] #include <bits/stdc++.h> using namespace std; const int N = 1e5; vector <int> G[N + 10]; int n,color[N+10],cnt[2]; void dfs(int x, i…

传送门:CF-862A A. Mahmoud and Ehab and the MEX time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Ol…

Mahmoud and Ehab and yet another xor task 存在的元素的方案数都是一样的, 啊, 我好菜啊. 离线之后用线性基取check存不存在,然后计算答案. #include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair…

862C - Mahmoud and Ehab and the xor 思路:找两对异或后等于(1<<17-1)的数(相当于加起来等于1<<17-1),两个再异或一下就变成0了,0异或x等于x.所以只要把剩下的异或起来变成x就可以了.如果剩下来有3个,那么,这3个数可以是x^i^j,i,j. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back…