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#include<bits/stdc++.h> using namespace std; const int N=500005; typedef long long ll; int T,n,k,t,h[N]; ll dp[N][2]; struct edge{   int ver,net;   ll val; }f[N*2]; void add(int x,int y,int z){   f[++t].net=h[x];   h[x]=t,f[t].ver=y;   f[t].val=z; } void dfs(int x,int fa){   dp[x][0]=dp[x][1]=0;   priority_queue<ll> q;   for(int i=h[x];i;i=f[i].net){     int y=f[i].ver;     if(y==fa) continue;     dfs(y,x);     dp[x][0]+=dp[y][0];     q.push(dp[y][1]+f[i].val-dp[y][0]);   }   dp[x][1]=dp[x][0];   int p=k;   while(p--&&q.size()&&q.top()>0){     if(p) dp[x][1]+=q.top();     dp[x][0]+=q.top();     q.pop();   }   } int main(){   scanf("%d",&T);   while(T--){     int x,y,z;t=0;     scanf("%d%d",&n,&k);     for(int i=1;i<n;++i){       scanf("%d%d%d",&x,&y,&z);       add(x,y,z),add(y,x,z);     }     dfs(1,0);     printf("%lld\n",dp[1][0]);     for(int i=1;i<=n;++i) h[i]=0,dp[i][0]=dp[i][1]=0;   }   return 0; } // Good edges✅ Share @coding_000❤️

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int dp[603]; int mex(vector<int>&v, int i, int j){      set<int>st;      for(int ind = i;ind <= j;ind++) st.insert(v[ind]);      int ans = 1;      for(auto pr : st){           if(pr == ans) ans++;      }      return ans; } int love(vector<int>& arr, int ind, int k){          int n = arr.size();     if(k == 1) return mex(arr, ind, n - 1);                                                 if(dp[ind] != -1)  return dp[ind];     int maxi = INT_MIN;     for(int i = ind;i <= (n - k);i++){        int ans = mex(arr, ind, i) + helper(arr, i + 1, k - 1);        maxi = max(maxi, ans);     }     return dp[ind] = maxi; } int solve(int N, int K, vector<int>A){      memset(dp, -1, sizeof(dp));     return love(A, 0 , K); } // Max Strikes againhttps://t.me/coding_000❤️

int dp[603]; int mex(vector<int>&v, int i, int j){      set<int>st;      for(int ind = i;ind <= j;ind++) st.insert(v[ind]);      int ans = 1;      for(auto pr : st){           if(pr == ans) ans++;      }      return ans; } int coding_000(vector<int>& arr, int ind, int k){          int n = arr.size();     if(k == 1) return mex(arr, ind, n - 1);                                                 if(dp[ind] != -1)  return dp[ind];     int maxi = INT_MIN;     for(int i = ind;i <= (n - k);i++){        int ans = mex(arr, ind, i) + helper(arr, i + 1, k - 1);        maxi = max(maxi, ans);     }     return dp[ind] = maxi; } int solve(int N, int K, vector<int>A){      memset(dp, -1, sizeof(dp));     return coding_000(A, 0 , K); } // Max Strikes again✅ Share @coding_000❤️

int solve(vector<int>& arr) {       if(arr.size()==1)     {         if(arr[0]==0) return 0;         else return 1;     } coding_000     int used = 0;     int start = 0;     int n = arr.size();     int result = 0;     for (int i = 0; i < n; i++)     {         while ((used&arr[i])!=0)         {             used ^= arr[start];             start++;         }         used |= arr[i];         if (start < i)             result+=i - start + 1;     }     return 2*result; } // Count subsets✅ Share @coding_000❤️

#include <iostream> #include <vector> #include <queue> #include <algorithm> using namespace std; const int INF = 1e9; struct Node {     int value, dist;     Node(int v, int d) : value(v), dist(d) {} }; int main() {     int n, m;     cin >> n >> m;     vector<int> A(n);     vector<vector<int>> graph(n, vector<int>());     vector<vector<int>> dist(n, vector<int>(n, INF));     for (int i = 0; i < n; ++i) {         cin >> A[i];     }     for (int i = 0; i < m; ++i) {         int x, y;         cin >> x >> y;         --x; --y; // Convert to 0-based indexing         graph[x].push_back(y);         graph[y].push_back(x);         dist[x][y] = dist[y][x] = 1;     }     // Floyd-Warshall algorithm to calculate shortest distances     for (int k = 0; k < n; ++k) {         for (int i = 0; i < n; ++i) {             for (int j = 0; j < n; ++j) {                 dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);             }         }     }     priority_queue<Node, vector<Node>, greater<Node>> pq;     for (int i = 0; i < n; ++i) {         if (A[i] > 0) {             pq.push(Node(i, 0));         }     }     long long cost = 0;     while (!pq.empty()) {         Node node = pq.top();         pq.pop();         int u = node.value;         int d = node.dist;         if (A[u] > 0) {             cost += d * A[u];             A[u] = 0;             for (int v : graph[u]) {                 if (A[v] > 0) {                     pq.push(Node(v, d + 1));                 }             }         }     }     for (int i = 0; i < n; ++i) {         if (A[i] > 0) {             cout << -n << endl;             return 0;         }     }     cout << cost << endl;     return 0; } Tree=0✅ All pass✅ Share @coding_000❤️ More Share More Solutions❤️👨‍💻

int min_operations(string s) {     int n = s.length();     vector<vector<int>> dp(n, vector<int>(n, INT_MAX));     for (int i = 0; i < n; i++) {         dp[i][i] = 0;     }     for (int len = 2; len <= n; len++) {         for (int i = 0; i <= n - len; i++) {             int j = i + len - 1;             if (s[i] == s[j]) {                 dp[i][j] = min(dp[i][j], dp[i + 1][j - 1]);             } else {                 for (int k = i; k < j; k++) {                     dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + 1);                 }             }         }     }     return dp[0][n - 1]; } // String dot✅ Share @coding_000❤️

#include <bits/stdc++.h> using namespace std; #define ll long long ll dp[10005][2]; const int mod = (1e9 + 7); ll solve(ll i, ll o, ll n, ll k, vector<ll> &v) {     if (i > n) {         return 0;     }     if (dp[i][o] != -1)         return dp[i][o];     ll ans = 0;     if (o == 0) {         for (int j = i; j < min(n + 1, i + k); j++) {             ans = max(ans, solve(j + 1, 1 - o, n, k, v)) % mod;         }     } else {         ll ta = 0;         ll mx = 0;         for (int j = i; j < min(n + 1, i + k); j++) {             if (ta + v[j] >= 0) {                 ta += v[j];                 mx = max(mx, ta);             } else {                 ta = 0;             }             ll c = solve(j + 1, 1 - o, n, k, v) % mod;             ans = max(ans, ((j - i + 1) * mx) % mod + c % mod);         }     }     return dp[i][o] = ans; } int main() {     ll n, k;     cin >> n >> k;     vector<ll> v;     ll neg = 0;     for (int i = 0; i < n; i++) {         ll x;         cin >> x;         v.push_back(x);         if (x <= 0)             neg++;     }     if (neg == v.size()) {         cout << 0 << endl;         return 1;     }     memset(dp, -1, sizeof(dp));       cout << max(solve(0, 0, v.size() - 1, k, v), solve(0, 1, v.size() - 1, k, v)) << endl; } // Array Segments✅✅ Share @coding_000

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#include <iostream> #include <vector> const int MOD = 1000000007; using namespace std; int countBeautifulSequences(int n) {     if (n == 1) {         return 1;     }     vector<int> dp(n + 1, 0);     dp[1] = 1;     for (int i = 1; i <= n; ++i) {         for (int j = i; j <= n; ++j) {             dp[j] = (dp[j] + dp[j - i]) % MOD;         }     }     return (dp[n] - 1 + MOD) % MOD; } int main() {     int n;     cin >> n;     int result = countBeautifulSequences(n);     cout << result << endl;     return 0; } Count Beautiful sequence✅ Share @coding_000❤️

#include <bits/stdc++.h> using namespace std; void solve(vector<vector<int>> vv, int operation, int xx, int yy, int &res) {     for (int i = 1; i < 3; i++)     {         int sum1 = 0;         for (int j = 0; j < vv.size(); j++)         {             sum1 += vv[j][i - 1];         }         int sum2 = 0;         for (int j = 0; j < vv.size(); j++)         {             sum2 += vv[j][i];         }         if (sum1 == sum2)         {             res = min(res, operation);         }         return;     }     for (int i = 0; i < vv.size(); i++)     {         solve(vv, operation, xx, yy, res);         vector<int> p1 = vv[i];         reverse(p1.begin(), p1.end());         solve(vv, operation + yy, xx, yy, res);         vector<int> p2 = vv[i];         int temp1 = p2[0];         int temp2 = p2[1];         int temp3 = p2[2];         p2[2] = temp1;         p2[1] = temp3;         p2[0] = temp2;         solve(vv, operation + xx, xx, yy, res);         vector<int> p3 = vv[i];         temp1 = p2[0];         temp2 = p2[1];         temp3 = p2[2];         p2[2] = temp2;         p2[1] = temp1;         p2[0] = temp3;         solve(vv, operation + xx, xx, yy, res);     }     return; } int main() {     ios_base::sync_with_stdio(false);     cin.tie(NULL);     int t = 1;     while (t--)     {         int n = 0, m = 0, a = 0, b = 0, c = 0, d = 0, sum = 0, diff = 0, maxN = 0, minN = 0, count = 0, temp = 0;         bool flag = false;         cin >> n;         cin >> m;         int xx;         cin >> xx;         int yy;         cin >> yy;         vector<vector<int>> vv(n, vector<int>(m));         for (int i = 0; i < n; i++)         {             for (int j = 0; j < m; j++)             {                 cin >> vv[i][j];             }         }         if (n == 1)         {             cout << -1 << endl;             continue;         }         int res = INT_MAX;         solve(vv, 0, xx, yy, res);         cout << res << endl;     }     return 0; } Pay for Gift ✅ Infosys share @coding_000❤️

#include <iostream> #include <vector> #include <deque> #include <unordered_map> using namespace std; int smartTaxiDriver(int N, int K, vector<int>& T, vector<int>& P, vector<int>& C) {     unordered_map<int, vector<pair<int, int>>> g;     for (int i = 0; i < N - 1; ++i) {         g[P[i] - 1].push_back({i + 2, C[i]});     }     int mpc = 0;     for (int sn = 1; sn <= N; ++sn) {         vector<int> d(N + 1, -1);         d[sn] = 0;         deque<pair<int, int>> q = {{sn, 0}};         while (!q.empty()) {             auto [cn, cd] = q.front();             q.pop_front();             for (auto& [ne, rd] : g[cn]) {                 if (d[ne] == -1 || d[ne] > cd + rd) {                     d[ne] = cd + rd;                     q.push_back({ne, d[ne]});                 }             }         }         int pc = 0;         for (int des : T) {             if (d[des] <= K) {                 pc++;             }         }         mpc = max(mpc, pc);     }     return mpc - 1; } int main() {     int N, K;     cin >> N >> K;     vector<int> T(N - 1), P(N - 1), C(N - 1);     for (int i = 0; i < N - 1; ++i) {         cin >> T[i];     }     for (int i = 0; i < N - 1; ++i) {         cin >> P[i];     }     for (int i = 0; i < N - 1; ++i) {         cin >> C[i];     }     int res = smartTaxiDriver(N, K, T, P, C);     cout << res << endl;     return 0; } Smart Taxi Driver ✅ Infosys @coding_000❤️