ACDream 1213 Matrix Multiplication (01矩阵处理)

Matrix Multiplication

Time Limit: 2000/1000MS (Java/Others) 

Memory Limit: 128000/64000KB (Java/Others)

Submit Statistic Next Problem

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;
}