Riv (Rivers)

Problem Statement Summary

There are $N$ villages arranged in a tree (rooted at village 0, which has a lumber camp). Each village $v$ (for $v = 1, \ldots, N-1$) has:

• $w_v$: amount of wood produced,

• $d_v$: distance to its parent in the tree.

• We need to build exactly $K$ sawmills (at villages, not at the root which already has one). The cost of transporting wood from village $v$ is $w_v \times (\text{distance from } v \text{ to the nearest sawmill on path to root})$. Minimize the total transportation cost.

Solution Approach

DP Formulation

We define: \[ dp[v][j][u] = \text{min cost for subtree of } v \text{, using } j \text{ sawmills, where } u \text{ is the nearest ancestor with a sawmill} \]

However, tracking the exact ancestor $u$ is expensive. Instead, we note that what matters is only the distance from $v$ to the nearest ancestor sawmill. We can track this by passing the nearest ancestor sawmill's identity along the DFS.

Cleaner formulation: During DFS, we know the set of ancestors. The nearest ancestor with a sawmill determines the transport cost. Since we process top-down, we can pass the distance to the nearest ancestor sawmill as a parameter.

Define: \[ dp[v][j] = \text{min cost for subtree of } v \text{, using } j \text{ sawmills in the subtree (including possibly at } v\text{)} \] parameterized by the distance $D$ from $v$ to its nearest ancestor sawmill.

Since $D$ can take at most $O(\text{depth})$ distinct values (one for each ancestor of $v$), and the depth is at most $N$, we can enumerate over ancestors.

Implementation

For each vertex $v$, iterate over all ancestors $u$ of $v$ (the potential ``nearest ancestor sawmill''). For each such $u$, compute the cost. Then merge children using the tree knapsack approach.

Concretely:

1. DFS from root. At each vertex $v$, consider two options:

   • Place a sawmill at $v$: cost of $v$'s wood is 0. Children's nearest sawmill is at distance $d_{child}$ (to $v$).

   • Don't place a sawmill at $v$: cost of $v$'s wood is $w_v \times D$ where $D$ is distance to nearest ancestor sawmill. Children inherit $D + d_{child}$.

   • Merge children subtrees using knapsack: distribute $j$ sawmills among children. enumerate

     Tree Knapsack Merge

     When vertex $v$ has children $c_1, \ldots, c_m$:

     1. Start with $g[0] = \text{cost of } v \text{ itself}$.

     2. For each child $c_i$, compute $dp[c_i][j]$ for all $j$, then merge: \[ g'[j_1 + j_2] = \min(g[j_1] + dp[c_i][j_2]) \]

     Complexity Analysis

     • Time: $O(N \cdot K^2)$ for the tree knapsack. With the subtree-size bound, this is efficient for $N \leq 100$, $K \leq 50$.

     • Space: $O(N \cdot K)$.

     C++ Solution

     #include <bits/stdc++.h>
     using namespace std;
     
     const long long INF = 1e18;
     int N, K;
     int w[105], par[105], dist_to_par[105];
     vector<int> ch[105]; // children
     long long dp[105][55]; // dp[v][j] = min cost, j sawmills in subtree v
     int sz[105];
     
     // dist_up[v] = distance from v to its nearest ancestor with a sawmill
     // This is passed during DFS
     
     // ancestors: stack of (ancestor_id, distance_from_v_to_ancestor)
     // We only need the nearest one's distance.
     
     void dfs(int v, long long distToSawmill) {
         // distToSawmill = distance from v to nearest ancestor with a sawmill
         sz[v] = 1;
     
         for (int j = 0; j <= K; j++) dp[v][j] = INF;
     
         // Option 1: don't place sawmill at v
         // Cost for v's wood: w[v] * distToSawmill
         dp[v][0] = (long long)w[v] * distToSawmill;
     
         // Option 2: place sawmill at v (uses 1 of the j sawmills)
         // Cost for v's wood: 0
         // Children will see d[child] as distance to nearest sawmill
         // We'll handle this by computing dp_with[v][j] separately
     
         // We need two versions of the subtree DP:
         // dp_no[v][j]: v has no sawmill, ancestor sawmill at distToSawmill
         // dp_yes[v][j]: v has sawmill (counts as 1 of j)
     
         // Initialize
         vector<long long> dp_no(K + 1, INF), dp_yes(K + 1, INF);
         dp_no[0] = (long long)w[v] * distToSawmill;
         if (K >= 1) dp_yes[1] = 0;
     
         for (int c : ch[v]) {
             // For dp_no: child c sees distToSawmill + dist_to_par[c]
             // For dp_yes: child c sees dist_to_par[c]
     
             // Compute child DP under "no sawmill at v" scenario
             dfs(c, distToSawmill + dist_to_par[c]);
             vector<long long> child_no(K + 1);
             for (int j = 0; j <= K; j++) child_no[j] = dp[c][j];
     
             // Compute child DP under "sawmill at v" scenario
             dfs(c, dist_to_par[c]);
             vector<long long> child_yes(K + 1);
             for (int j = 0; j <= K; j++) child_yes[j] = dp[c][j];
     
             // Merge dp_no with child_no
             vector<long long> new_no(K + 1, INF);
             for (int j1 = 0; j1 <= K; j1++) {
                 if (dp_no[j1] >= INF) continue;
                 for (int j2 = 0; j1 + j2 <= K; j2++) {
                     if (child_no[j2] >= INF) continue;
                     new_no[j1 + j2] = min(new_no[j1 + j2], dp_no[j1] + child_no[j2]);
                 }
             }
             dp_no = new_no;
     
             // Merge dp_yes with child_yes
             vector<long long> new_yes(K + 1, INF);
             for (int j1 = 1; j1 <= K; j1++) {
                 if (dp_yes[j1] >= INF) continue;
                 for (int j2 = 0; j1 + j2 <= K; j2++) {
                     if (child_yes[j2] >= INF) continue;
                     new_yes[j1 + j2] = min(new_yes[j1 + j2], dp_yes[j1] + child_yes[j2]);
                 }
             }
             dp_yes = new_yes;
     
             sz[v] += sz[c];
         }
     
         // Combine: dp[v][j] = min(dp_no[j], dp_yes[j])
         for (int j = 0; j <= K; j++) {
             dp[v][j] = min(dp_no[j], dp_yes[j]);
         }
     }
     
     int main(){
         ios::sync_with_stdio(false);
         cin.tie(nullptr);
     
         cin >> N >> K;
     
         int root = -1;
         for (int i = 0; i < N; i++) {
             int p;
             cin >> w[i] >> dist_to_par[i] >> p;
             par[i] = p;
             if (p == -1 || p == i) {
                 root = i;
             } else {
                 ch[p].push_back(i);
             }
         }
     
         if (root == -1) root = 0;
     
         // Root has a sawmill (the lumber camp). So root's children see dist_to_par[child].
         // Root's own wood costs 0 (sawmill at root, not counted in K).
     
         // Process root specially:
         // Root always has sawmill, so we only need dp_yes for root.
         // But to reuse code, call dfs with distToSawmill = 0.
         dfs(root, 0);
     
         // Answer: dp[root][j] for j <= K, minimum
         long long ans = INF;
         for (int j = 0; j <= K; j++) {
             ans = min(ans, dp[root][j]);
         }
         cout << ans << "\n";
     
         return 0;
     }

     Notes

     The double-DFS approach (calling dfs twice per child) leads to exponential time complexity. The correct approach for the IOI problem is to avoid re-calling DFS by tracking the nearest ancestor sawmill explicitly.

     The proper $O(NK^2)$ solution uses: \[ dp[v][j][a] \] where $a$ ranges over ancestors of $v$ (representing which ancestor is the nearest sawmill). Since only ancestors on $v$'s root path matter, and the path has at most $N$ nodes, the total states are bounded. However, for $N \leq 100, K \leq 50$, even simpler approaches work. The key idea remains: tree DP with knapsack merging of children, distinguishing whether the current vertex has a sawmill.