SGU - 196 - Matrix Multiplication （矩阵乘法）

196. Matrix Multiplication

time limit per test: 0.25 sec.
memory limit per test: 65536 KB

input: standard
output: standard

Let us consider an undirected graph G = <V, E> which has N vertices and M edges. Incidence matrix of this graph is an N × M matrix A = {a _ij}, such that a _ij 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 where A ^T is A transposed, i.e. an M × N matrix obtained from A by turning its columns to rows and vice versa. 

Input

The first line of the input file contains two integer numbers — N and M (2 le N le 10,000, 1 le M le 100,000). 2M 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 test(s)

Input

4 4 1 2 1 3 2 3 2 4 

Output

18 

[submit]

[forum]

Author:	Andrew Stankevich, Georgiy Korneev
Resource:	Petrozavodsk Winter Trainings 2003
Date:	2003-02-06

思路：自己手动执行几下就可以发现规律，这里存储图中边的方法为完全关联矩阵（离散数学中有介绍），而本题的规律为只要统计每个定点的出现次数的平方即可

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
using namespace std;

int num[10005];
int N, M;

int main() {
	while(scanf("%d %d", &N, &M) != EOF) {
		memset(num, 0, sizeof(num));
		for(int i = 0; i < M; i++) {
			int a, b;
			scanf("%d %d", &a, &b);
			num[a]++, num[b]++;
		}
		LL ans = 0;
		for(int i = 1; i <= N; i++) {
			ans += (num[i] * num[i]);
		}
			
		cout << ans << endl;
	}
	return 0;
}