Create Travelling_Salesman_Problem.cpp

Author: rishi058

C++/Travelling_Salesman_Problem.cpp

@@ -0,0 +1,119 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+#define int long long
+#define F(a,b,i) for (int i = a; i < b; i++)
+#define all(v) v.begin(), v.end()
+
+typedef long double ld;
+typedef vector<int> vi;
+
+//------------------------------------------------------
+
+/*
+Travelling Salesman Problem...
+
+A weighted Graph is Given, you have to traverse through each vertex exactly once and return to the
+starting vertex such that the cost of travelling is least. (Also Known as Hamiltonian Cycle..)
+
+Brute Force -->
+Find the all premutation of the Paths and choose min cost path by iteratin them all 
+T.C --> O(n!)
+
+D.P -->  ( Top - Down )
+Take 1 node as root node.
+
+Traverse through the node and build rec tree(it will have 2^n vertices)...
+
+From bottom of tree calculate the min cost req to reach par node...
+
+int mask = bit-masking [1->0001(represent A is vis), 3->(0011)...] is used to keep track of visited nodes while building rec tree..
+
+I have taken the max_constraint for 2d matrix but-->
+
+DP size = [2^n][n]     [ row->keep track of visitness || col ->keep track of nodes ]
+
+T.C --> O(n^2 * 2^n) ..
+
+*/
+
+
+/*
+
+**********************************************************************************
+|   Better Aprroach is to use map , instead of 2D matrix for memoization..      |
+|   as u can see the output --> most of the dp entries in unutilized...         |
+**********************************************************************************
+
+*/
+
+
+int dist[10001][10001] ; // Adjacency Matrix of Graph..
+
+int dp[10001][10001] ;
+
+int VISITED_ALL, n;
+
+int TSP(int mask, int pos){
+    if(mask==VISITED_ALL){
+        return dist[pos][0]; // agar sabhi node visit kar liya to uss node se direct root node(i.e 1) tak ka cost add kr k return...
+    }
+
+    if(dp[mask][pos]!=-1){
+        return dp[mask][pos];
+    }
+
+    int ans = INT_MAX;
+
+    F(0,n,city){
+        if((mask&(1<<city))==0){             //[chcking no city is visited more than once, as 1&1=1 always]
+            int newAns = dist[pos][city] + TSP( mask|(1<<city), city);  // or operation is used to mark the visted city ..
+            ans = min(ans,newAns);
+        }
+
+    }
+
+    return dp[mask][pos] = ans;
+
+}
+
+
+int32_t main()
+{
+    cin>>n;
+
+    memset(dp, -1, sizeof(dp));
+
+    F(0,n,i){
+        F(0,n,j){
+            cin>>dist[i][j];
+        }
+    }
+
+    VISITED_ALL = (1<<n) - 1;
+
+    cout<<TSP(1,0)<<" "<<dp[1][0]<<"\n";
+
+    F(0,16,i){
+        F(0,4,j){
+            cout<<dp[i][j]<<" ";
+        }
+        cout<<"\n";
+    }
+
+}
+
+/*
+gfg e.g->
+4
+0 10 15 20
+10 0 25 25
+15 25 0 30
+20 25 30 0
+4
+0 20 42 25
+20 0 30 34
+42 30 0 10
+25 34 10 0
+
+*/