### Problem Statement :

```Determine the minimum cost to provide library access to all citizens of HackerLand. There are n cities numbered from 1 to n. Currently there are no libraries and the cities are not connected. Bidirectional roads may be built between any city pair listed in cities. A citizen has access to a library if:

1. Their city contains a library.
2. They can travel by road from their city to a city containing a library.

unction Description

Complete the function roadsAndLibraries in the editor below.

int n: integer, the number of cities
int c_lib: integer, the cost to build a library
int cities[m][2]: each cities[ i ] contains two integers that represent cities that can be connected by a new road
Returns
- int: the minimal cost

Input Format

The first line contains a single integer q, that denotes the number of queries.

The subsequent lines describe each query in the following format:
- The first line contains four space-separated integers that describe the respective values of n , m , c_lib and c_road, the number of cities, number of roads, cost of a library and cost of a road.
- Each of the next m lines contains two space-separated integers, u[ i ] and v[ i ] , that describe a bidirectional road that can be built to connect cities c[ i ] and v[ i ].

Sample Input

STDIN       Function
-----       --------
2           q = 2
3 3 2 1     n = 3, cities[] size m = 3, c_lib = 2, c_road = 1
1 2         cities = [[1, 2], [3, 1], [2, 3]]
3 1
2 3
6 6 2 5     n = 6, cities[] size m = 6, c_lib = 2, c_road = 5
1 3         cities = [[1, 3], [3, 4],...]
3 4
2 4
1 2
2 3
5 6

Sample Output

4
12```

### Solution :

```                            ```Solution in C :

#include <stdlib.h>
#include <stdio.h>
#include <assert.h>

#define swap_(x, y) { int z = x; x = y; y = z; }

typedef long long ll;

typedef struct L
{
int *xs;
int n;
int size;
} L;

{
if (l->n == l->size)
{
l->size *= 2;
l->xs = realloc(l->xs, sizeof(int) * l->size);
assert(l->xs);
}
l->xs[l->n++] = x;
}

void ini(L *l)
{
l->n = 0;
l->size = 4;
l->xs = malloc(sizeof(int) * l->size);
assert(l->xs);
}

L *create()
{
L *l = malloc(sizeof(L));
assert(l);
ini(l);
return l;
}

L *ls[100000];

ll solve()
{
int n, m;
ll rc, lc;
scanf("%d%d%lld%lld", &n, &m, &lc, &rc);
for (int i = 0; i < n; ++i)
{
ls[i] = NULL;
}
int count = 0;
for (int _i = 0; _i < m; ++_i)
{
int i, j;
scanf("%d%d", &i, &j);
i--; j--;
if (lc <= rc)
{
continue;
}
if (ls[i] == ls[j])
{
if (ls[i] == NULL)
{
L *l = create();
count++;
ls[i] = l;
ls[j] = l;
}
}
else
{
count++;
if (ls[i] == NULL)
{
swap_(i, j);
}
if (ls[j] == NULL)
{
ls[j] = ls[i];
}
else
{
if (ls[i]->n < ls[j]->n)
{
swap_(i, j);
}
L *l = ls[j];
for (int p = 0; p < l->n; ++p)
{
int k = l->xs[p];
ls[k] = ls[i];
}
free(l->xs);
free(l);
}
}
}
return rc * count + lc * (n - count);
}

int main()
{
int q;
scanf("%d", &q);
for (int i = 0; i < q; ++i)
{
ll min_cost = solve();
printf("%lld\n", min_cost);
}
return 0;
}```
```

```                        ```Solution in C++ :

In   C++ :

#include <bits/stdc++.h>
using namespace std;
vector<int> p;
int f(int a){return p[a]==a?a:p[a]=f(p[a]);}
void u(int a, int b){p[f(a)] = f(b);}
signed main()
{
int T;cin >> T;while(T--){
int N, M, a, b;
long long c, d;
cin >> N >> M >> c >> d;
p.clear();p.resize(N);
iota(p.begin(), p.end(), 0);
while(M--){
cin >> a >> b;
--a, --b;
u(a, b);
}
int comp=0;
for(int i=0;i<N;++i){
if(p[i]==i)++comp;
}
cout << (comp*c+(N-comp)*min(c, d)) << "\n";
}

return 0;
}```
```

```                        ```Solution in Java :

In   C++ :

#include <bits/stdc++.h>
using namespace std;
vector<int> p;
int f(int a){return p[a]==a?a:p[a]=f(p[a]);}
void u(int a, int b){p[f(a)] = f(b);}
signed main()
{
int T;cin >> T;while(T--){
int N, M, a, b;
long long c, d;
cin >> N >> M >> c >> d;
p.clear();p.resize(N);
iota(p.begin(), p.end(), 0);
while(M--){
cin >> a >> b;
--a, --b;
u(a, b);
}
int comp=0;
for(int i=0;i<N;++i){
if(p[i]==i)++comp;
}
cout << (comp*c+(N-comp)*min(c, d)) << "\n";
}

return 0;
}```
```

```                        ```Solution in Python :

In   Python3 :

#!/bin/python3

import sys

def find_root(roots, city):
i = roots[city]
while(i != roots[i]):
i = roots[i]
return roots[i]

q = int(input().strip())
for a0 in range(q):
n,m,x,y = input().strip().split(' ')
n,m,x,y = [int(n),int(m),int(x),int(y)]

root = [x for x in range(n+1)]
cost = 0

if(x <= y):
print(n*x)
for _ in range(m):
x = input()
continue
for a1 in range(m):
city_1,city_2 = input().strip().split(' ')
city_1,city_2 = [int(city_1),int(city_2)]
temp1 = find_root(root, city_1)
temp2 = find_root(root, city_2)
if(temp1 != temp2):
root[temp1] = temp2
cost += y
for i in range(1,n+1):
if(i == root[i]):
cost +=x
print(cost)```
```

## Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

## Tree: Preorder Traversal

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## Tree: Postorder Traversal

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