ZOJ 2316 Matrix Multiplication

Matrix Multiplication

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on ZJU. Original ID: 2316
64-bit integer IO format: %lld      Java class name: Main

 

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 = {aij}, such that aij 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.

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

18

 

Source

Andrew Stankevich's Contest #1

 

解题：题意转化后，就是计算途中有多少条，长度为2的路径。注意是无向图。。。。每个顶点，看它的度是多少。从这些度里面选取2个的组合数。。。。。。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 10010;
int d[maxn];
int main() {
    int t,u,v,n,m,ans;
    scanf("%d",&t);
    while(t--){
        scanf("%d %d",&n,&m);
        memset(d,0,sizeof(d));
        for(int i = 0; i < m; i++){
            scanf("%d %d",&u,&v);
            ++d[u];
            ++d[v];
        }
        ans = 0;
        for(int i = 1; i <= n; i++)
            ans += d[i]*(d[i]-1)/2;
        ans = (ans + m)<<1;
        printf("%d
",ans);
        if(t) puts("");
    }
    return 0;
}