3 Steps(二分图)

C - 3 Steps

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Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

Rng has a connected undirected graph with N vertices. Currently, there are M edges in the graph, and the i-th edge connects Vertices Ai and Bi.

Rng will add new edges to the graph by repeating the following operation:

• Operation: Choose u and v (u≠v) such that Vertex v can be reached by traversing exactly three edges from Vertex u, and add an edge connecting Vertices u and v. It is not allowed to add an edge if there is already an edge connecting Vertices u and v.

Find the maximum possible number of edges that can be added.

Constraints

• 2≤N≤105
• 1≤M≤105
• 1≤Ai,Bi≤N
• The graph has no self-loops or multiple edges.
• The graph is connected.

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Input

Input is given from Standard Input in the following format:

N M
A1 B1
A2 B2
:
AM BM

Output

Find the maximum possible number of edges that can be added.

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Sample Input 1

Copy

6 5
1 2
2 3
3 4
4 5
5 6

Sample Output 1

Copy

4

If we add edges as shown below, four edges can be added, and no more.

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Sample Input 2

Copy

5 5
1 2
2 3
3 1
5 4
5 1

Sample Output 2

Copy

5

Five edges can be added, for example, as follows:

• Add an edge connecting Vertex 5 and Vertex 3.
• Add an edge connecting Vertex 5 and Vertex 2.
• Add an edge connecting Vertex 4 and Vertex 1.
• Add an edge connecting Vertex 4 and Vertex 2.
• Add an edge connecting Vertex 4 and Vertex 3.

//Atcoder的题目还是有新意啊，可以收获不少

题意： n 个点 m 条边， 组成一个无向连通图，重复操作， 如果 a 点到 b 点距离为 3 ，并且没有连回 a ，就添加一条 a - b 的边。没有自环，问最多能添加几条边。

分析可知，如有图有奇数环，必然可加成完全图

如果图是二分图，则会变成完全二分图，

否则最终变为完全图

二分图dfs染色即可

 #include <bits/stdc++.h>
 using namespace std;
 # define LL          long long
 # define pr          pair
 # define mkp         make_pair
 # define lowbit(x)   ((x)&(-x))
 # define PI          acos(-1.0)
 # define INF         0x3f3f3f3f3f3f3f3f
 # define eps         1e-
 # define MOD         

 inline int scan() {
     int x=,f=; char ch=getchar();
     while(ch<''||ch>''){if(ch=='-') f=-; ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}
     return x*f;
 }
 inline void Out(int a) {
     if(a<) {putchar('-'); a=-a;}
     if(a>=) Out(a/);
     putchar(a%+'');
 }
 # define MX
 /**************************/
 int n,m;
 vector<int> G[MX];
 int clo[MX];

 bool check()
 {
     bool ok=;
     for (int i=;i<=n;i++)
     {
         if (G[i].size()>=)
         {
             if (ok) return ;
             ok=;
         }
     }
     return ;
 }

 int dfs(int p,int s,int pre)
 {
     clo[p] = s%+;
     for (int i=;i<G[p].size();i++)
     {
         int v = G[p][i];
         if (v==pre) continue;
         if (!clo[v])
         {
             if (!dfs(v,s+,p))
                 return ;
         }
         else if(clo[v]==clo[p]) return ;
     }
     return ;
 }

 int bipartite()
 {
     memset(clo,,sizeof(clo));
     if (!dfs(,,-)) return ;
     return ;
 }

 int main()
 {
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         for (int i=;i<=n;i++) G[i].clear();

         for (int i=;i<=m;i++)
         {
             int x,y;
             scanf("%d%d",&x,&y);
             G[x].push_back(y);
             G[y].push_back(x);
         }
         if (!check())
             printf("0\n");
         else if (!bipartite())
             printf("%lld\n",(LL)n*(n-)/-m);
         else
         {
             LL b=,w=;
             for (int i=;i<=n;i++)
             {
                 if (clo[i]==) b++;
                 else w++;
             }
             printf("%lld\n",b*w-m);
         }
     }
     return ;
 }