Fully_Connected_4_Nodes.cpp

// [In the near future, human engineers have realized that the optimal model for 
// organizing infrastructure in cities is the fully connected graph. 
// Kura recently moved into the town of Koenigsberg. Its a small town that consists 
// of only 4 houses, for each of its inhabitants.

// As the new arrival, Kura discovers that he has three neighbors: Leon, Lieb and Hamil. 
// Also, he finds a map of the town in your house. It looks like this in 
// a weighted graph format: Kura's house is marked as h, Leon's house is marked 
// as a, Lieb's house is marked as b, and Hamil's house is marked as c and 
// ha, hb, hc, ab, bc & ca represent the repective distances between the houses. 

// From the map, Kura was able to determine that the distance from any place ' x '
// on the map to a place ' y ' on the map is same as the distance from ' y ' to ' x '. 
// After spending the day unpacking, Kura decided to visit 2 of his neighbors. 
// Since, he has limited energy to walk, you must help Kura determine 
// the minimum distance he must walk to meet any 2 of his neighbors and come back home. 
// He doesn't mind visiting the same neighbor or passing the same road multiple times. 
// The only goal is to minimize the total distance traveled.


// Input Format:
// -------------
// space separated 6 integers representing ha, hb, hc, ac,ab,bc  

// Output Format:
// --------------
// A single number representing minimum distance that kura needs to travel to 
// visit any 2 of his neighbors and return home. 

// Constraints:
// ------------
// 1<=ha,hb,hc,ac,ab,bc<= 10^12

// Sample input:
// -------------
// 3 2 8 8 1 4

// Sample output: 
// --------------
// 6



#include<bits/stdc++.h>
using namespace std;
int main(){
    int n;
    unordered_map<string,int> m;
    vector<int> v(n,0);
    for(int i=0;i<n;i++){
        cin>>v[i];
    }
    m["ha"]=v[0];
    m["hb"]=v[1];
    m["hc"]=v[2];
    m["ac"]=v[3];
    m["ab"]=v[4];
    m["bc"]=v[5];
    int mina=m["ha"]+m["ab"]+m["hb"];
    int minb=m["ha"]+m["ac"]+m["hc"];
    int minc=m["hc"]+m["bc"]+m["hc"];
    cout<<min(mina,min(minb,minc));
}