Codeforces - 862B Mahmoud and Ehab and the bipartiteness (二分图)
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 such a way, that for each edge (u, v) that belongs to the graph, u and v belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.
Dr. Evil gave Mahmoud and Ehab a tree consisting of n nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?
A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same .
The first line of input contains an integer n — the number of nodes in the tree (1 ≤ n ≤ 105).
The next n - 1 lines contain integers u and v (1 ≤ u, v ≤ n, u ≠ v) — the description of the edges of the tree.
It's guaranteed that the given graph is a tree.
Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions.
3 1 2 1 3
0
5 1 2 2 3 3 4 4 5
2
Tree definition: https://en.wikipedia.org/wiki/Tree_(graph_theory)
Bipartite graph definition: https://en.wikipedia.org/wiki/Bipartite_graph
In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0.
In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5).
[二分图]
两个集合, 每个集合内部不存在边边关系,
[思路]
可以用dfs搜索, 用 flag 标记 flag=1-> A集合 flag=0 ->B集合, 把所有有边边关系的进入不同两个集合中,
flag不断翻转
[代码实现]
#include <bits/stdc++.h>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <deque>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x,b,n) lower_bound(b+1,b+1+n,x)-b
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define SHUT ios_base::sync_with_stdio(false); cout.setf(ios::fixed);cout.precision(20); cout.tie(nullptr); cin.tie(nullptr);
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r
#define FI(n) IO::read(n)
#define Be IO::begin()
using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e7+5;
const int MAXN=1e5+5;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};
struct node{
int v,u,next;
}edge[maxn];
int cot,head[maxn],vis[maxn];
int n,m;
void init()
{
cot=0;
mem(vis,0);
mem(head,-1);
}
void add(int u,int v)
{
edge[++cot].v=v;
edge[cot].u=u;
edge[cot].next=head[u];
head[u]=cot;
}
set<int> A,B;
void dfs(int u,int father,int flag)
{
flag^=1;
if(vis[u])
return ;
vis[u]=1;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(v!=father)
{
// cout<<"flag is "<<flag<<" "<<v<<endl;
if(flag)
A.insert(v);
else
B.insert(v);
dfs(v,u,flag);
}
}
}
int main()
{
cin>>n;
init();
int x,y;
for(int i=1;i<=n-1;i++)
{
cin>>x>>y;
add(x,y);
add(y,x);
}
for(int i=1;i<=n;i++)
{
if(i==1)
A.insert(i);
dfs(i,0,1);
}
ll sumA=A.size();
ll sumB=B.size();
//cout<<sumA<< " "<<sumB<<endl;
ll ans=sumA*sumB-n+1;
cout<<ans<<endl;
return 0;
}