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。。。。。。

 

 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 }

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