ACdream 1213 Matrix Multiplication

Matrix Multiplication

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

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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 41 21 32 32 4

Sample Output

18

Source

Andrew Stankevich Contest 1

Manager

mathlover

题意：给一个无向图，设这个图的关联矩阵为A。然后计算ATA这个矩阵中，所有元素的和。

关于关联矩阵，百度百科有详细介绍：http://baike.baidu.com/link?url=H9T8Sm0NTuoKwBiKS7nOBFn5oV10M0_UW0XhXAYA2mDjIhq2PUr84sknOaYpYmZocFXDYdp-Kij-RKIIP9rDN_

思路：观察关联矩阵，对于每条边的两个顶点，这条边对结果的贡献是这两个顶点的度，所以只需要扫描每条边，然后加上出现的顶点的度就可以了。

#include <cstring>#include <cstdio>typedef long long LL;const int mx=100010;int edge[mx][2];int cnt[mx];int main() {    int n,m;    while(~scanf("%d%d",&n,&m)) {        memset(cnt,0,sizeof(cnt));        for(int i=1; i<=m; i++) {            int u,v;            scanf("%d%d",&u,&v);            edge[i][0]=u;            edge[i][1]=v;            cnt[u]++;            cnt[v]++;        }        LL ans=0;        for(int i=1; i<=m; i++) {            ans+=cnt[edge[i][0]];            ans+=cnt[edge[i][1]];        }        printf("%lld\n",ans);    }    return 0;}