Matrix Multiplication
Time Limit: 2000/1000MS (Java/Others)
Memory Limit: 128000/64000KB (Java/Others)
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 = {ai,j}, such that ai,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 ATA.
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 4 1 2 1 3 2 3 2 4
Sample Output
18
Source
Andrew Stankevich Contest 1
题意：给出一个N个点个M条无向边的关系构成一个01矩阵，求01矩阵A*A^T后新矩阵的每个元素的值。
分析：这个邻接矩阵的转置矩阵还是这个矩阵本身，那么题目就变成了求A^2，有1W个点的话，用普通的矩阵肯定存不下来的，会超内存。
设原矩阵为A，矩阵相乘之后为B，转置之后矩阵还是本身，按照矩阵乘法本来是A的第 i 行乘第 j 列得到 Aij， 转变思路思考一下，可以变成第 Aij 可以变成每一行和该行本身相乘。
可以先计算出每一列的1的个数用一维数组C存下来，然后如果Bij=1的话，就Bij*C[i]，可以纸上模拟一下。
换种说法，就是先存下每个点的度，然后把所有边的两边的点的度相加。
数据可能会超int，用long long
#pragma comprint(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<time.h>
#include<algorithm>
#define LL __int64
#define FIN freopen("in.txt","r",stdin)
using namespace std;
const int MAXN=10000+5;
int cnt[MAXN];
struct node
{
int x,y;
}edge[100000+5];
int main()
{
int n,m;
while(scanf("%d %d",&n,&m)!=EOF)
{
memset(cnt,0,sizeof(cnt));
for(int i=0;i<m;i++)
{
int u,v;
scanf("%d %d",&u,&v);
cnt[u]++;
cnt[v]++;
edge[i].x=u;
edge[i].y=v;
}
long long ans=0;
for(int i=0;i<m;i++)
{
int u=edge[i].x;
int v=edge[i].y;
ans+=cnt[u];
ans+=cnt[v];
}
printf("%lld\n",ans);
}
return 0;
}