ZOJ - 2316 Matrix Multiplication

最新推荐文章于 2020-04-13 01:47:39 发布

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本文探讨了一个图论问题，即如何通过构建图的邻接矩阵并计算该矩阵与其转置的乘积来解决特定求和问题。文章提供了一个简洁高效的算法实现，并附带示例输入输出以帮助理解。

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Let us consider undirected graph G = which has N vertices and M edges. Incidence matrix of this graph is N * M matrix A = {a_ij}, such that a _ij 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 A^TA.

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. 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

1

4 4
1 2
1 3
2 3
2 4

Sample Output

题意：就是n个点，m条边，建成一个矩阵，问这个矩阵和他的转置矩阵的乘积。

思路：例如样例，建成的矩阵是 0 1 1 0

1 0 1 1

1 1 0 0

0 1 0 0

乘以它的转置相当于乘它本身，a[1][2]这个数要和a[2][1],a[2][3],a[2][4]相乘得3个1，a[3][2],a[4][2]同理

每个矩阵位置为1的都可以这样计算，得出规律了

#nclude<iostream>

#include<cstdio>
#include<cstring>
using namespace std;
int main(){
    int a[10005];
    int t,n,m;
    int x,y,i;
    scanf("%d",&t);
    while(t--){
        memset(a,0,sizeof(a));
        scanf("%d%d",&n,&m);
        while(m--){
            scanf("%d%d",&x,&y);
            a[x]++;
            a[y]++;
        }
        int sum=0;
        for(i=1;i<=n;i++)
            sum+=a[i]*a[i];
        if(t==0)
            printf("%d\n",sum);
        else
            printf("%d\n\n",sum);
    }
    return 0;
}