D - Make Bipartite 2

https://atcoder.jp/contests/abc282/tasks/abc282_d

Simple Graph

https://mathworld.wolfram.com/SimpleGraph.html

 

A simple graph, also called a strict graph (Tutte 1998, p. 2), is an unweighted, undirected graph containing no graph loops or multiple edges (Gibbons 1985, p. 2; West 2000, p. 2; Bronshtein and Semendyayev 2004, p. 346). A simple graph may be either connected or disconnected.

Unless stated otherwise, the unqualified term "graph" usually refers to a simple graph. A simple graph with multiple edges is sometimes called a multigraph (Skiena 1990, p. 89).

Multigraph

https://mathworld.wolfram.com/Multigraph.html

 

The term multigraph refers to a graph in which multiple edges between nodes are either permitted (Harary 1994, p. 10; Gross and Yellen 1999, p. 4) or required (Skiena 1990, p. 89, Pemmaraju and Skiena 2003, p. 198; Zwillinger 2003, p. 220). West (2000, p. xiv) recommends avoiding the term altogether on the grounds of this ambiguity.

Bipartite Graph -- 二分图

https://mathworld.wolfram.com/BipartiteGraph.html

 

A bipartite graph, also called a bigraph, is a set of graph vertices decomposed into two disjoint sets such that no two graph vertices within the same set are adjacent. A bipartite graph is a special case of a k-partite graph with . The illustration above shows some bipartite graphs, with vertices in each graph colored based on to which of the two disjoint sets they belong.

Bipartite graphs are equivalent to two-colorable graphs. All acyclic graphs are bipartite. A cyclic graph is bipartite iff all its cycles are of even length (Skiena 1990, p. 213).

二分图判断

code

https://www.geeksforgeeks.org/bipartite-graph/

https://www.zhihu.com/question/292465499

如何判定

相信大家一定对二分图有了一个初步的了解，那么我们该如何判定二分图呢？

首先我们要引进一个概念，染色

 

判断二分图的常见方法是染色法：用两种颜色，对所有顶点逐个染色，且相邻顶点染不同的颜色

如果发现相邻顶点染了同一种颜色，就认为此图不为二分图

当所有顶点都被染色，且没有发现同色的相邻顶点，就退出

参考

https://atcoder.jp/contests/abc282/submissions/37362401

需要处理存在多个孤立图的情况。

对于每个独立图， 分别计算 同色的点的数量，

独立图之间任意连线是有效的。

#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
using ll = long long;
using P = pair<int,int>;
using Graph = vector<vector<int>>;
using mint = modint1000000007;

map<ll,vector<ll>> uv;
vector<ll> node;
vector<ll> black;
vector<ll> white;
bool color(int current, int col){
  node[current] = col;
  if(col == 1){
    black.push_back(current);
  }
  else{
    white.push_back(current);
  }
  ll rev;
  if(col == 1){
    rev = -1;
  }
  else{
    rev = 1;
  }
  for(ll nod: uv[current]){
    if(node[nod] == col){
      return false;
    }
    else if(node[nod] == 0){
      if (!color(nod, rev)){
        return false;
      }
    }
  }
  return true;
}

int main() {
  ll n,m;
  cin >> n >> m;
  node.assign(n+1,0);
  rep(i,0,m){
    int u,v;
    cin >> u >> v;
    uv[u].push_back(v);
    uv[v].push_back(u);
  }
  ll sameColors = 0;
  rep(i,1,n+1){
    if(node[i] == 0){
      black.clear();
      white.clear();
      if(!color(i,1)){
        cout << "0" << endl;
        return 0;
      }
      ll bSize = (int)black.size();
      ll wSize = (int)white.size();
      sameColors += bSize*(bSize-1)/2 + wSize*(wSize-1)/2;
    }
  }
  //同じ色
  cout << n*(n-1)/2-m-sameColors << endl;
  return 0;
}