Create Shortest path in a graph with exactly K edges.cpp

Author: nikolatechie

Shortest path in a graph with exactly K edges.cpp

@@ -0,0 +1,96 @@
+#include <bits/stdc++.h>
+#include <ext/rope>
+using namespace std;
+using namespace __gnu_cxx;
+
+using ci = const int;
+using ld = long double;
+using llint = long long;
+using ullint = unsigned long long;
+using pii = pair <int,int>;
+using pcc = pair <char,char>;
+using pss = pair <string,string>;
+using vi = vector <int>;
+using vb = vector <bool>;
+using vii = vi::iterator;
+
+#define INF (1<<30)
+#define MOD 1000000007
+#define mp make_pair
+#define mt make_tuple
+#define all(c) c.begin(), c.end()
+#define ms(name,val) memset(name, val, sizeof name)
+#define np nullptr
+
+
+const int MX = 100;
+int n, DP[MX][MX][MX], g[MX][MX];
+
+int shortestPath(int a, int b, int k)
+{
+    for (int len = 0; len <= k; ++len)
+    {
+        for (int src = 0; src < n; ++src)
+        {
+            for (int dest = 0; dest < n; ++dest)
+            {
+                // initialize value
+                DP[src][dest][len] = INF;
+
+                // 1st base case
+                if (!len && src == dest)
+                    DP[src][dest][0] = 0;
+
+                // 2nd base case
+                if (len == 1 && g[src][dest] != INF)
+                    DP[src][dest][1] = g[src][dest];
+
+                if (len < 2) continue;
+                // now solve for general cases
+
+                // adjacents of src
+                for (int adj = 0; adj < n; ++adj)
+                {
+                    if (adj != src && adj != dest &&
+                        g[src][adj] != INF && DP[adj][dest][len-1] != INF)
+                    {
+                        DP[src][dest][len] = min(DP[src][dest][len],
+                                                 g[src][adj] + DP[adj][dest][len-1]);
+                    }
+                }
+            }
+        }
+    }
+
+    return DP[a][b][k];
+}
+
+int main()
+{
+    ios_base::sync_with_stdio(0);
+    //cin.tie(0);
+
+    int m;
+    cin >> n >> m;
+
+    for (int t1 = 0; t1 < n; ++t1)
+    {
+        for (int t2 = 0; t2 < n; ++t2)
+            g[t1][t2] = INF;
+
+        g[t1][t1] = 0;
+    }
+
+    while (m)
+    {
+        --m;
+        int a, b, c;
+        cin >> a >> b >> c;
+        g[a][b] = c;
+    }
+
+    cout << shortestPath(0, n-1, n>>1) << '\n';
+
+
+    return 0;
+}