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ACDream 1213 Matrix Multiplication (01矩阵处理)

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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 = {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

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

ACDream 1213 Matrix Multiplication (01矩阵处理)

#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; }

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