ZOJ 2316 Matrix Multiplication（找规律）（矩阵和它的转置矩阵之积）

Matrix Multiplication

Let us consider undirected graph G = which has N vertices and M edges. Incidence matrix of this graph is N * M matrix A = {aij}, such that aij 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 ATA.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input

The first line of the input file contains two integer numbers - N and M (2 <= N <= 10 000, 1 <= M <= 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 Input

1

4 4
1 2
1 3
2 3
2 4

Sample Output

18


思路：要求的是A矩阵和A的转置矩阵的积，列一下矩阵会发现，实际上就是求Aij（1<=i<=n,1<=j<=m）的平方的和（这好像是求一个矩阵和它的转置矩阵之积的规律？）。
其实根据线性代数里面的公式，就是矩阵的第K行的和和第K列的和的乘积之和。

代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define maxn 100010int a[maxn];int n,m;int main(){    int t;    scanf("%d",&t);    while(t--)    {        memset(a,0,sizeof(a));        int x,y;        scanf("%d%d",&n,&m);        for(int i=1;i<=m;++i)        {            scanf("%d%d",&x,&y);            ++a[x],++a[y];        }        int ans=0;        for(int i=1;i<=n;++i)            ans+=a[i]*a[i];        printf("%d\n",ans);        if(t)            printf("\n");    }    return 0;}