Let us consider undirected graph G = which has N vertices and M edges. Incidence matrixof 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 edgeand 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. Alledges 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
題解：T組數據，n,m，然後m行邊，根據m行邊轉換成一個n行m列的矩陣，求這個矩陣的轉置矩陣乘以此矩陣的出來的矩陣的元素之和……
比賽的時候一看題目，矩陣相乘，一臉懵逼，不會，趕緊找了本線代書看了下，看了半個小時終於懂了，但一看數據範圍，我了個大草，這怎麼存圖，怎麼運算，肯定妥妥的雙超……
賽後看了下別人的的代碼……我心中一萬隻國寶奔騰而過……不過還是沒弄懂什麼意思，如果哪位大神知道，請拯救一下迷途的羔羊……
#include<bits/stdc++.h>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<string>
#include<math.h>
#include<map>
#include<queue>
#include<stack>
#define INF 0x3f3f3f3f
#define ll long long
#define For(i,a,b) for(int i=a;i<b;i++)
#define sf(a) scanf("%d",&a)
#define sfs(a) scanf("%s",a)
#define sff(a,b) scanf("%d%d",&a,&b)
#define sfff(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define pf(a) printf("%d\n",a)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int x[10005];
int main()
{
int t,n,m,falg=0;
sf(t);
while(t--)
{
if(falg)printf("\n");falg=1;
mem(x,0);
sff(n,m);
int a,b;
For(i,0,m)
{
sff(a,b);
x[a]++;x[b]++;
}
ll s=0;
For(i,1,n+1)
{
s+=x[i]*x[i];
}
cout<<s<<endl;
}
}