3 Steps(二分图)

C - 3 Steps

---

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.

---

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.

---

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.

---

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-8
# define MOD         1000000007
 
inline int scan() {
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
inline void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
# define MX 100005
/**************************/
int n,m;
vector<int> G[MX];
int clo[MX];
 
bool check()
{
    bool ok=0;
    for (int i=1;i<=n;i++)
    {
        if (G[i].size()>=2)
        {
            if (ok) return 1;
            ok=1;
        }
    }
    return 0;
}
 
int dfs(int p,int s,int pre)
{
    clo[p] = s%2+1;
    for (int i=0;i<G[p].size();i++)
    {
        int v = G[p][i];
        if (v==pre) continue;
        if (!clo[v])
        {
            if (!dfs(v,s+1,p))
                return 0;
        }
        else if(clo[v]==clo[p]) return 0;
    }
    return 1;
}
 
int bipartite()
{
    memset(clo,0,sizeof(clo));
    if (!dfs(1,1,-1)) return 0;
    return 1;
}
 
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for (int i=1;i<=n;i++) G[i].clear();
 
        for (int i=1;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
");
        else if (!bipartite())
            printf("%lld
",(LL)n*(n-1)/2-m);
        else
        {
            LL b=0,w=0;
            for (int i=1;i<=n;i++)
            {
                if (clo[i]==1) b++;
                else w++;
            }
            printf("%lld
",b*w-m);
        }
    }
    return 0;
}