Acdream 1213——Matrix Multiplication

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 41 21 32 32 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


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;}