2017 JUST Programming Contest 3.0 B. Linear Algebra Test_. linear algebra test time limit per test3.0 s-优快云博客

本文链接：https://blog.youkuaiyun.com/Fire_to_cheat_/article/details/78173660

本文介绍了一种快速计算矩阵配对可乘性的方法，通过使用哈希映射（map），将时间复杂度降低到O(n)。文章提供了一个具体的实现示例，并解释了如何处理大规模输入输出。

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Dr. Wail is preparing for today's test in linear algebra course. The test's subject is Matrices Multiplication.

Dr. Wail has n matrices, such that the size of the ith matrix is (ai × bi), where ai is the number of rows in the ith matrix, and bi is the number of columns in the ith matrix.

Dr. Wail wants to count how many pairs of indices i and j exist, such that he can multiply the ith matrix with the jth matrix.

Dr. Wail can multiply the ith matrix with the jth matrix, if the number of columns in the ith matrix is equal to the number of rows in the jthmatrix.

Input

The first line contains an integer T (1 ≤ T ≤ 100), where T is the number of test cases.

The first line of each test case contains an integer n (1 ≤ n ≤ 105), where n is the number of matrices Dr. Wail has.

Then n lines follow, each line contains two integers ai and bi (1 ≤ ai, bi ≤ 109) (ai ≠ bi), where ai is the number of rows in the ith matrix, and bi is the number of columns in the ith matrix.

Output

For each test case, print a single line containing how many pairs of indices i and j exist, such that Dr. Wail can multiply the ith matrix with the jth matrix.

Example

input
1
5
2 3
2 3
4 2
3 5
9 4

output
5

Note

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printfinstead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

In the first test case, Dr. Wail can multiply the 1st matrix (2 × 3) with the 4th matrix (3 × 5), the 2nd matrix (2 × 3) with the 4th matrix (3 × 5), the 3rd matrix (4 × 2) with the 1st and second matrices (2 × 3), and the 5th matrix (9 × 4) with the 3rd matrix (4 × 2). So, the answer is 5.

题意：给你多个矩阵ai表示行数，bi表示列数

要你求行数与列数相等的对数有多少种？

思路：利用map可以把复杂度降到O（n）

需要注意的是

2 3

3 2

这一类的情况结果是2对，而不是一对；另外由于数据是1e9，所以以后见到1e9都用上long long

#include <iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<map>
using namespace std;
const int maxn=1e5+10;
typedef long long ll;
struct node
{
    ll col,row;
    int flag;
};
struct mnode
{
    int flag;
};


int main()
{
    ll t;
    scanf("%lld",&t);
    while(t--)
    {
        map<ll,node> m;
       // map<mnode,mnode>mab;
       map<int,int>mab;
        ll n;
        scanf("%lld",&n);
        ll a,b;
        ll ans=0,flag=0;
        for(ll i=1;i<=n;i++)
            {
                scanf("%lld%lld",&a,&b);
                mab[a]=b;
                    m[a].col+=1;
                    m[b].row+=1;

            }
        map<ll,node>::iterator it;
        for(it=m.begin();it!=m.end();it++){
                ans+=it->second.col*it->second.row;

        }
        printf("%lld\n",ans-flag);
    }
    return 0;
}