Add a lot more problems and a TODO list

Educational Codeforces Round 3/E. Minimum spanning tree for each edge.cpp

@@ -0,0 +1,191 @@
+/* Problem URL: https://codeforces.com/contest/609/problem/E */
+
+#include <bits/stdc++.h>
+#include <ext/pb_ds/assoc_container.hpp>
+#include <ext/pb_ds/tree_policy.hpp>
+
+using namespace std;
+using namespace __gnu_pbds;
+
+template <class T, class comp = less<>>
+using ordered_set = tree<T, null_type , comp , rb_tree_tag , tree_order_statistics_node_update>;
+
+#define V vector
+
+#define rmin(a, b) a = min(a, b)
+#define rmax(a, b) a = max(a, b)
+
+#define rep(i, lim) for (int i = 0; i < (lim); i++)
+#define nrep(i, s, lim) for (int i = s; i < (lim); i++)
+
+#define repv(i, v) for (auto &i : (v))
+#define fillv(v) for (auto &itr_ : (v)) { cin >> itr_; }
+#define sortv(v) sort(v.begin(), v.end())
+#define all(v) (v).begin(), (v).end()
+
+using vi = vector<int>;
+using vvi = vector<vi>;
+using vvvi = vector<vvi>;
+using vvvvi = vector<vvvi>;
+
+using ll = long long;
+
+using vl = vector<ll>;
+using vvl = vector<vl>;
+using vvvl = vector<vvl>;
+using vvvvl = vector<vvvl>;
+
+template<class v>
+auto operator<<(ostream &os, const vector<v> &vec)->ostream& {
+    os << vec[0];
+    for (size_t i = 1; i < vec.size(); i++) {
+        os << ' ' << vec[i];
+    }
+    os << '\n';
+    return os;
+}
+
+template<class v>
+auto operator>>(istream &is, vector<v> &vec)->istream& {
+    for (auto &i : vec) {
+        is >> i;
+    }
+    return is;
+}
+
+template<class v>
+auto operator<<(ostream &os, const vector<vector<v>> &vec)->ostream& {
+    for (auto &i : vec) {
+        os << i[0];
+        for (size_t j = 1; j < i.size(); j++) {
+            os << ' ' << i[j];
+        }
+        os << '\n';
+    }
+    return os;
+}
+
+template<class v>
+auto operator>>(istream  &is, vector<vector<v>> &vec)->istream& {
+    for (auto &i : vec) {
+        for (auto &j : i) {
+            is >> j;
+        }
+    }
+    return is;
+}
+
+int main()
+{
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int n, m;
+    cin >> n >> m;
+
+    V<tuple<ll, int, int, int>> edges(m);
+    rep(i, m) {
+        auto &[c, u, v, j] = edges[i];
+        cin >> u >> v >> c;
+        u--, v--;
+        j = i;
+    }
+
+    sortv(edges);
+
+    vi dsu(n);
+    rep(i, n) {
+        dsu[i] = i;
+    }
+
+    function<int(int)> find_p = [&](int i) {
+        if (dsu[i] == i) {
+            return i;
+        }
+        return dsu[i] = find_p(dsu[i]);
+    };
+
+    auto join = [&](int a, int b) {
+        a = find_p(a);
+        b = find_p(b);
+        dsu[b] = a;
+        return a != b;
+    };
+
+    ll ans = 0;
+    V<V<pair<int, ll>>> graph(n);
+
+    for (auto [c, u, v, j] : edges) {
+        if (join(u, v)) {
+            ans += c;
+            graph[u].emplace_back(v, c);
+            graph[v].emplace_back(u, c);
+        }
+    }
+
+    vvi parent(20, vi(n));
+    vvl maximal(20, vl(n));
+    vi depth(n);
+
+    function<void(int, int, ll)> dfs = [&](int i, int p, ll c) {
+        parent[0][i] = p;
+        maximal[0][i] = c;
+        depth[i] = depth[p] + 1;
+
+        nrep(j, 1, 20) {
+            parent[j][i] = parent[j - 1][parent[j - 1][i]];
+            maximal[j][i] = max(maximal[j - 1][i], maximal[j - 1][parent[j - 1][i]]);
+        }
+
+        for (auto [j, c] : graph[i]) {
+            if (j == p) {
+                continue;
+            }
+
+            dfs(j, i, c);
+        }
+    };
+
+    dfs(0, 0, 0);
+
+    auto lca = [&](int a, int b) {
+        if (depth[a] > depth[b]) {
+            swap(a, b);
+        }
+
+        ll ans = 0;
+        int diff = depth[b] - depth[a];
+
+        rep(i, 20) {
+            if ((diff >> i) & 1) {
+                rmax(ans, maximal[i][b]);
+                b = parent[i][b];
+            }
+        }
+
+        if (a == b) {
+            return ans;
+        }
+
+        for (int i = 19; i >= 0; i--) {
+            if (parent[i][a] != parent[i][b]) {
+                rmax(ans, maximal[i][a]);
+                rmax(ans, maximal[i][b]);
+                a = parent[i][a];
+                b = parent[i][b];
+            }
+        }
+
+        return max({ans, maximal[0][a], maximal[0][b]});
+    };
+
+    vl act(m);
+
+    for (auto [c, u, v, i] : edges) {
+        act[i] = ans - lca(u, v) + c;
+    }
+
+    repv(i, act) {
+        cout << i << '\n';
+    }
+}