3 Steps

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.

解析：判断是否是二分图，用深搜遍历，二分图的判断是用染色法，从而得出相应的答案；
代码：

#include<iostream>
#include<cstdio>
#include<vector> 
#include<cstring>
using namespace std;
const int maxn = 1e5 + 5;
vector<int> list[maxn];
int vis[maxn];
int lose=0;
void bfs(int ne,int be,int re)
{
    if(!lose)
    {
        if(vis[ne]==-1)
            vis[ne]=re;
        else
            if(vis[ne]!=re)
            {
                lose=1;
                return;
            }
            else
                return;
        for(int i=0;i<list[ne].size();i++)
        {
            if(list[ne][i]!=be)
                bfs(list[ne][i],ne,!re);    
        }

    }

}
int main()
{
    long long m,n;
    scanf("%lld%lld",&m,&n);
    for(int i=0;i<n;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        list[x].push_back(y);
        list[y].push_back(x);
    }
    memset(vis,-1,sizeof(vis));
    bfs(1,0,1);
    long long q=0,p=0;
    for(int i=1;i<=m;i++)
    {
        if(vis[i]==1)
            q++;
        else
            p++;
    }
    if(lose)
        printf("%lld\n",m*(m-1)/2-n);
    else
        printf("%lld\n",q*p-n);

    return 0;
}