Problem: There are N towns on a plane. The i-th town is located at the coordinates (x_i,y_i). There may be more than one town at the same coordinates.  You can build a road between two towns at coordinates (a,b) and (c,d) for a cost of min(|a-c|,|b-d|) yen (the currency of Japan). It is not possible to build other types of roads.  Your objective is to build roads so that it will be possible to travel between every pair of towns by traversing roads. At least how much money is necessary to achieve this?

Submission attempts (in order):



class uf_tree():

    def __init__(self,n):
        self.n = n
        self.par = [i for i in range(n)]

    def indep(self):
        dep =0
        for i in range(self.n):
            if i == self.par[i]:
                dep += 1

        return dep

    def unite(self,child,mas):
        if self.root(child) == self.root(mas):
            return
        else:
            self.par[self.root(child)] = self.root(mas)

    def root(self,i):
        if self.par[i] == i:
            return i
        else:
            self.par[i] = self.root(self.par[i])
            return self.par[i]

    def same(self,x,y):
        if self.root(x) == self.root(y):
            return True
        else:
            return False
def kruscal(n,edge_arr):
    edge_arr.sort(key=lambda x:x[2])
    ans = []
    tree = uf_tree(n)
    for e in edge_arr:
        s = e[0]-1
        g = e[1]-1
        if tree.same(s,g):
            continue
        else:
            ans.append(e)
            tree.unite(s,g)
    return ans

def main():
    n = int(input())
    p = [[0,0,0] for i in range(n)]
    edge = []
    for i in range(n):
        x,y = map(int,input().split())
        p[i][0] = x
        p[i][1] = y
        p[i][2] = i+1

    p.sort(key = lambda x:x[0])
    for i in range(n-1):
        cost = min(abs(p[i+1][0]-p[i][0]),abs(p[i+1][1]-p[i][1]))
        edge.append([p[i][2],p[i+1][2],cost])

    p.sort(key = lambda x:x[1])
    for i in range(n-1):
        cost = min(abs(p[i+1][0]-p[i][0]),abs(p[i+1][1]-p[i][1]))
        edge.append([p[i][2],p[i+1][2],cost])

    ans = kruscal(n,edge)
    tot = 0
    for e in ans:
        tot += e[2]
    print(tot)


class uf_tree():

    def __init__(self,n):
        self.n = n
        self.par = [i for i in range(n)]
        self.rank = [0 for i in range(n)]

    def indep(self):
        dep =0
        for i in range(self.n):
            if i == self.par[i]:
                dep += 1

        return dep

    def unite(self,x,y):
        if self.root(x) == self.root(y):
            return
        else:
            rootx = self.root(x)
            rooty = self.root(y)
            if self.rank[rootx] < self.rank[rooty]:
                self.par[rootx] = rooty
            else:
                self.par[rooty] = rootx
                if self.rank[rootx] == self.rank[rooty]:
                    self.rank[rootx] += 1
    def root(self,i):
        if self.par[i] == i:
            return i
        else:
            self.par[i] = self.root(self.par[i])
            return self.par[i]

    def same(self,x,y):
        return self.root(x) == self.root(y)

def kruscal(n,edge_arr):
    edge_arr.sort(key=lambda x:x[2])
    ans = []
    tree = uf_tree(n)
    for e in edge_arr:
        s = e[0]-1
        g = e[1]-1
        if tree.same(s,g):
            continue
        else:
            ans.append(e)
            tree.unite(s,g)
    return ans

p = []
edge = []
n = int(input())
for i in range(n):
    x,y = map(int,input().split())
    p.append([x,y,i+1])

p.sort(key = lambda x:x[0])
for i in range(n-1):
    cost = min(abs(p[i+1][0]-p[i][0]),abs(p[i+1][1]-p[i][1]))
    edge.append([p[i][2],p[i+1][2],cost])

p.sort(key = lambda x:x[1])
for i in range(n-1):
    cost = min(abs(p[i+1][0]-p[i][0]),abs(p[i+1][1]-p[i][1]))
    edge.append([p[i][2],p[i+1][2],cost])

ans = kruscal(n,edge)
tot = 0
for e in ans:
    tot += e[2]
print(tot)

Verdicts: Wrong Answer

Accepted