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]

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Author:Andrew Stankevich, Georgiy KorneevResource:Petrozavodsk Winter Trainings 2003Date: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 longusing 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;}