How to create an unordered_map of pairs in C++?

Unordered Map does not contain a hash function for a pair like it has for int, string, etc, So if we want to hash a pair then we have to explicitly provide it with a hash function that can hash a pair. unordered_map can takes upto 5 arguments: 

• Key : Type of key values
• Value : Type of value to be stored against the key
• Hash Function : A function which is used to hash the given key. If not provided it uses default hash function.
• Pred : A function which is used so that no two keys can have same hash values
• Alloc : An object used to define the memory model for the map

hash_function can be anything, given that it can hash the given key.
Prerequisite : How to create an unordered_map of user defined class?

CPP

// CPP program to demonstrate implementation of
// unordered_map for a pair.
#include <bits/stdc++.h>
using namespace std;

// A hash function used to hash a pair of any kind
struct hash_pair {
    template <class T1, class T2>
    size_t operator()(const pair<T1, T2>& p) const
    {
        auto hash1 = hash<T1>{}(p.first);
        auto hash2 = hash<T2>{}(p.second);

        if (hash1 != hash2) {
            return hash1 ^ hash2;
        }

        // If hash1 == hash2, their XOR is zero.
          return hash1;
    }
};

int main()
{
    // Sending the hash function as a third argument
    unordered_map<pair<int, int>, bool, hash_pair> um;

    // Creating some pairs to be used as keys
    pair<int, int> p1(1000, 2000);
    pair<int, int> p2(2000, 3000);
    pair<int, int> p3(3005, 3005);
    pair<int, int> p4(4000, 4000);

    // Inserting values in the unordered_map.
    um[p1] = true;
    um[p2] = false;
    um[p3] = true;
    um[p4] = false;

    cout << "Contents of the unordered_map : \n";
    for (auto p : um)
        cout << "[" << (p.first).first << ", "
             << (p.first).second << "] ==> " << p.second
             << "\n";

    return 0;
}

Output

Contents of the unordered_map : 
[4000, 4000] ==> 0
[3005, 3005] ==> 1
[1000, 2000] ==> 1
[2000, 3000] ==> 0

Note : We can create map for a pair. Check out map of pair. The reason is, map is based on self balancing BSTs and does not require a hash function.
Exercise Problem : Nikhil is a travelling salesman and today he is visiting houses in a new locality to sell encyclopedias. The new city is in the form of a grid of x*y(1<=x<=10^9, 1<=y<=10^9) and at every intersection there is a house. Now he is not very good with remembering the houses that he has already visited, so whenever he goes into a house he tells you the coordinate of the house. Your job is to remember the coordinate and at the end of the day tell him all the houses that he visited on that day.
Examples:

Input :
Enter the number of houses that he visited today :5 
Enter the coordinate of HouseNo. 1 :1000 12985 
Enter the coordinate of HouseNo. 2 :12548 25621 
Enter the coordinate of HouseNo. 3 :14586 26481 
Enter the coordinate of HouseNo. 4 :12 63 
Enter the coordinate of HouseNo. 5 :14689 36945 
Output : 
Houses that he visited today: 
12 63 
14689 36945 
14586 26481 
1000 12985 
12548 25621