AtCoder

Problem Statement

 

In the State of Takahashi in AtCoderian Federation, there are N cities, numbered 1,2,…,N. M bidirectional roads connect these cities. The i-th road connects City A_i and City B_i. Every road connects two distinct cities. Also, for any two cities, there is at most one road that directly connects them.

One day, it was decided that the State of Takahashi would be divided into two states, Taka and Hashi. After the division, each city in Takahashi would belong to either Taka or Hashi. It is acceptable for all the cities to belong Taka, or for all the cities to belong Hashi. Here, the following condition should be satisfied:

• Any two cities in the same state, Taka or Hashi, are directly connected by a road.

Find the minimum possible number of roads whose endpoint cities belong to the same state. If it is impossible to divide the cities into Taka and Hashi so that the condition is satisfied, print -1.

Constraints

 

• 2≤N≤700
• 0≤M≤N(N−1)⁄2
• 1≤A_i≤N
• 1≤B_i≤N
• A_i≠B_i
• If i≠j, at least one of the following holds: A_i≠A_j and B_i≠B_j.
• If i≠j, at least one of the following holds: A_i≠B_j and B_i≠A_j.

Input

 

Input is given from Standard Input in the following format:

N M
A1 B1
A2 B2
:
AM BM

Output

 

Print the answer.

Sample Input 1

 

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

Sample Output 1

 

4

For example, if the cities 1,2 belong to Taka and the cities 3,4,5 belong to Hashi, the condition is satisfied. Here, the number of roads whose endpoint cities belong to the same state, is 4.

Sample Input 2

 

5 1
1 2

Sample Output 2

 

-1

In this sample, the condition cannot be satisfied regardless of which cities belong to each state.

Sample Input 3

 

4 3
1 2
1 3
2 3

Sample Output 3

 

3

Sample Input 4

 

10 39
7 2
7 1
5 6
5 8
9 10
2 8
8 7
3 10
10 1
8 10
2 3
7 4
3 9
4 10
3 4
6 1
6 7
9 5
9 7
6 9
9 4
4 6
7 5
8 3
2 5
9 2
10 7
8 6
8 9
7 3
5 3
4 5
6 3
2 10
5 10
4 2
6 2
8 4
10 6

Sample Output 4

 

21


    打ARC的时候都想到模型了。。。但就是没做出来mmp，还是水平差啊QWQ
    主要没想到的地方是： 如果把补图二分图染色(不是二分图就无解)，每个联通分量挑一类点出来弄到一起一定能凑成最后的一个团。
    我也不知道为什么当时没想到QWQ，明明这么简单。。。。
    然后直接背包完了更新答案就好了QWQ，怎么看都是一个NOIP题，药丸药丸

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=705;

inline int Get(int x){ return x*(x-1)>>1;}

int num[3],n,m,ans=1<<30,uu,vv,col[N],now;
bool f[N][N],g[N][N];

bool dfs(int x,int c){
	col[x]=c,num[c]++;
	
	for(int i=1;i<=n;i++) if(!g[x][i])
	    if(!col[i]){ if(!dfs(i,3-c)) return 0;}
	    else if(col[i]==c) return 0;
	
	return 1;
}

int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	    scanf("%d%d",&uu,&vv),g[uu][vv]=g[vv][uu]=1;
	
	f[0][0]=1;
	for(int i=1;i<=n;i++) g[i][i]=1;
	
	for(int i=1;i<=n;i++) if(!col[i]){
		num[1]=num[2]=0;
		if(!dfs(i,1)){ puts("-1"); return 0;}
		
		now++;
		for(int j=0;j<=n;j++) if(f[now-1][j]) f[now][j+num[1]]=f[now][j+num[2]]=1;
	}
	
	for(int i=0;i<=n;i++) if(f[now][i]) ans=min(ans,Get(i)+Get(n-i));
	
	printf("%d
",ans);
	return 0;
}