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。。。。。。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 #include <climits>
7 #include <vector>
8 #include <queue>
9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 10010;
18 int d[maxn];
19 int main() {
20 int t,u,v,n,m,ans;
21 scanf("%d",&t);
22 while(t--){
23 scanf("%d %d",&n,&m);
24 memset(d,0,sizeof(d));
25 for(int i = 0; i < m; i++){
26 scanf("%d %d",&u,&v);
27 ++d[u];
28 ++d[v];
29 }
30 ans = 0;
31 for(int i = 1; i <= n; i++)
32 ans += d[i]*(d[i]-1)/2;
33 ans = (ans + m)<<1;
34 printf("%d\n",ans);
35 if(t) puts("");
36 }
37 return 0;
38 }