Algorithm
Problem link- https://www.spoj.com/problems/FRNDCIRC/
FRNDCIRC - FRIEND CIRCLE
Lucy has made too many friends but she does not know how many friends are in her circle. Assume that every relation is mutual. If Lucy is Patty's friend, then Patty is also Lucy's friend. Your task is to help Lucy in keeping track of each person's circle size.
Input Specification
The first line of input contains one integer T (T<=10) specifying the number of test cases to follow. Each test case begins with a line containing an integer N (N<=100000), the number of new relations. Each of the following N lines contains couple of strings denoting the names of two people who have just formed relation, separated by a space. Names will have no more than 20 characters.
Output Specification
Print a line containing one integer, the number of people in the combined circle of two people who have just become friends.
Input
1 4 Lucy Patty Patty Alice Alice Mira Tiffany Jayden
Output
2 3 4 2
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MOD 1000000007LL
#define EPS 1e-9
#define io ios_base::sync_with_stdio(false);cin.tie(NULL);
#define M_PI 3.14159265358979323846
template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}
const int MAXN = 2e5+5;
int n;
int parent[MAXN];
int size[MAXN];
void init(){
for(int i = 1;i <= 2*n; i++){
parent[i] = i;
size[i] = 1;
}
}
int root(int i){
while(i != parent[i]){
parent[i] = parent[parent[i]];
i = parent[i];
}
return i;
}
bool find(int a, int b){
return root(a) == root(b);
}
void union1(int a, int b){
int root_a = root(a);
int root_b = root(b);
if(root_a == root_b)
return;
if(size[root_a] < size[root_b]){
parent[root_a] = root_b;
size[root_b] += size[root_a];
}else{
parent[root_b] = root_a;
size[root_a] += size[root_b];
}
}
int main(){
io;
int t;
cin >> t;
while(t--){
cin >> n;
init();
unordered_map<string, int> M;
int val = 1;
for(int i = 0;i < n; i++){
string s1, s2;
cin >> s1 >> s2;
int a, b;
if(M.find(s1) == M.end()){
M[s1] = val;
a = val;
val++;
}else{
a = M[s1];
}
if(M.find(s2) == M.end()){
M[s2] = val;
b = val;
val++;
}else{
b = M[s2];
}
union1(a, b);
cout << size[root(a)] << endl;
}
}
return 0;
} Input
4
Lucy Patty
Patty Alice
Alice Mira
Tiffany Jayden
Output
3
4
2
Demonstration
SPOJ Solution-FRIEND CIRCLE-Solution in C, C++, Java, Python