Dynamic Programming III: Advanced

7. Practice Problems

1. Assign N tasks to N workers (cost matrix) minimising total cost — bitmask DP.
   N জন worker ও N tasks-এর cost matrix দেওয়া; minimum total cost বের করুন।
   Show Answer

   assign.cpp

   #include <bits/stdc++.h>
   using namespace std;
   int main() {
       int n = 4;
       int c[4][4] = {{9,2,7,8},{6,4,3,7},{5,8,1,8},{7,6,9,4}};
       vector<int> dp(1 << n, INT_MAX);
       dp[0] = 0;
       for (int mask = 0; mask < (1 << n); ++mask) {
           int i = __builtin_popcount(mask);
           if (i == n || dp[mask] == INT_MAX) continue;
           for (int j = 0; j < n; ++j) if (!(mask & (1 << j)))
               dp[mask | (1 << j)] = min(dp[mask | (1 << j)], dp[mask] + c[i][j]);
       }
       cout << "Min cost = " << dp[(1 << n) - 1] << "\n";
   }

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2. Count integers in [L, R] whose digit sum equals K (digit DP).
   [L, R] range-এ এমন কতগুলো integer যাদের digit sum = K।
   Show Answer

   digsum.cpp

   #include <bits/stdc++.h>
   using namespace std;
   string S; int K;
   long long memo[20][200][2]; bool seen[20][200][2];
   long long f(int p, int s, bool t) {
       if (p == (int)S.size()) return s == K;
       if (seen[p][s][t]) return memo[p][s][t];
       seen[p][s][t] = 1;
       int hi = t ? S[p] - '0' : 9;
       long long r = 0;
       for (int d = 0; d <= hi; ++d) if (s + d <= K)
           r += f(p + 1, s + d, t && (d == hi));
       return memo[p][s][t] = r;
   }
   long long cnt(long long N) {
       if (N < 0) return 0;
       S = to_string(N); memset(seen, 0, sizeof seen);
       return f(0, 0, true);
   }
   int main() {
       K = 10;
       cout << cnt(10000) - cnt(99) << "\n";
   }

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3. Diameter of a tree using two DFS or via tree DP (longest path in two subtrees).
   Tree-এর diameter (দীর্ঘতম পথ) tree DP দিয়ে।
   Show Answer

   diameter.cpp

   #include <bits/stdc++.h>
   using namespace std;
   vector<int> g[100005];
   int ans = 0;
   int dfs(int u, int par) {
       int m1 = 0, m2 = 0;
       for (int c : g[u]) if (c != par) {
           int d = dfs(c, u) + 1;
           if (d > m1) { m2 = m1; m1 = d; } else if (d > m2) m2 = d;
       }
       ans = max(ans, m1 + m2);
       return m1;
   }
   int main() {
       int e[][2] = {{1,2},{2,3},{2,4},{4,5},{5,6}};
       for (auto& x : e) { g[x[0]].push_back(x[1]); g[x[1]].push_back(x[0]); }
       dfs(1, 0);
       cout << "Diameter = " << ans << "\n";
   }

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4. Partition a set of n ≤ 16 integers into k subsets of equal sum (bitmask DP).
   n ≤ ১৬ সংখ্যা k সমান-যোগ subset-এ ভাগ করা যাবে কি না।
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   State: dp[mask] = remainder of (sum of picked elements) % target if we can fill complete subsets so far, else −1. Try adding each element to the current bucket.

   kparts.cpp

   #include <bits/stdc++.h>
   using namespace std;
   int main() {
       vector<int> a = {4,3,2,3,5,2,1}; int k = 4;
       int n = a.size(), tot = accumulate(a.begin(), a.end(), 0);
       if (tot % k) { cout << "NO\n"; return 0; }
       int tgt = tot / k;
       vector<int> dp(1 << n, -1);
       dp[0] = 0;
       for (int m = 0; m < (1 << n); ++m) if (dp[m] != -1)
           for (int i = 0; i < n; ++i) if (!(m & (1 << i)) && dp[m] + a[i] <= tgt)
               dp[m | (1 << i)] = (dp[m] + a[i]) % tgt;
       cout << (dp[(1 << n) - 1] == 0 ? "YES" : "NO") << "\n";
   }

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5. Rerooting: for every node compute the sum of distances to all other nodes.
   প্রতিটি node-এর জন্য বাকি সব node-এর সাথে দূরত্বের যোগফল।
   Show Answer

   reroot.cpp

   #include <bits/stdc++.h>
   using namespace std;
   const int N = 100005;
   vector<int> g[N]; long long sub[N], ans[N]; int sz[N], n;
   void dfs1(int u, int p) {
       sz[u] = 1; sub[u] = 0;
       for (int c : g[u]) if (c != p) {
           dfs1(c, u); sz[u] += sz[c]; sub[u] += sub[c] + sz[c];
       }
   }
   void dfs2(int u, int p) {
       for (int c : g[u]) if (c != p) {
           ans[c] = ans[u] - sz[c] + (n - sz[c]);
           dfs2(c, u);
       }
   }
   int main() {
       n = 5;
       int e[][2] = {{1,2},{1,3},{3,4},{3,5}};
       for (auto& x : e) { g[x[0]].push_back(x[1]); g[x[1]].push_back(x[0]); }
       dfs1(1, 0);
       ans[1] = sub[1];
       dfs2(1, 0);
       for (int i = 1; i <= n; ++i) cout << "ans[" << i << "]=" << ans[i] << " ";
       cout << "\n";
   }

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6. Hamiltonian path existence in a directed graph (n ≤ 18) — bitmask DP.
   n ≤ ১৮-এর directed graph-এ Hamiltonian path আছে কি না।
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   Use dp[mask][last] = true iff visited set is mask ending at last; transition over edges last → nxt. Answer = OR over dp[FULL][last].

7. Count numbers in [L, R] that have at least one digit equal to D — digit DP.
   [L, R] range-এ এমন সংখ্যা কতটি যাদের কমপক্ষে একটি digit D-এর সমান।
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   Easier: count those with no digit equal to D and subtract from R−L+1. State: (pos, tight, started). Use complementary counting.