Matrix Multiplication


Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)


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Problem Description


      Let us consider undirected graph G = {V; E} which has N vertices and M edges. Incidence matrix of this graph is N × M matrix A = {a i,j}, such that a i,j is 1 if i-th vertex is one of the ends of j -th edge and 0 in the other case. Your task is to find the sum of all elements of the matrix A TA.


Input


      The first line of the input file contains two integer numbers — N and M (2 ≤ N ≤ 10 000, 1 ≤ M ≤100 000). Then 2*M integer numbers follow, forming M pairs, each pair describes one edge of the graph. All edges are different and there are no loops (i.e. edge ends are distinct).


Output


      Output the only number — the sum requested.


Sample Input


4 4
1 2
1 3
2 3
2 4


Sample Output


18


Source


Andrew Stankevich Contest 1


Manager



mathlover




显然矩阵太大无法存， 所以要找规律，这题特殊在矩阵是关联矩阵，看图


Acdream 1213——Matrix Multiplication_Java



所以，C[i][j]可以看成是第k行里任意两个元素相乘的和


由于这里的矩阵是01矩阵，所以可以只考虑1 * 1，显然，第k行的和就是点k的度


比如第k行为 1 0 1 1，任意2个元素相乘的和是9，deg[k]^2 == 9 


#include<map>    
#include<set>    
#include<list>    
#include<stack>    
#include<queue>    
#include<vector>    
#include<cmath>    
#include<cstdio>    
#include<cstring>    
#include<iostream>    
#include<algorithm>    
    
using namespace std;

int deg[10010];

int main()
{
	int n, m;
	while (~scanf("%d%d", &n, &m))
	{
		memset( deg, 0, sizeof(deg) );
		int u, v;
		for (int i = 1; i <= m; i++)
		{
			scanf("%d%d", &u, &v);
			deg[u]++;
			deg[v]++;
		}
		long long ans = 0;
		for (int i = 1; i <= n; i++)
		{
			ans += (long long)(deg[i] * deg[i]);
		}
		printf("%lld\n", ans);
	}
	return 0;
}