From f8e979f1149aef6889a1ea1f830f7112d5bb06ed Mon Sep 17 00:00:00 2001
From: subcrip <contact@subc.rip>
Date: Wed, 29 May 2024 19:17:38 +0800
Subject: [PATCH] Add dp/knapsack.cc

Signed-off-by: subcrip <contact@subc.rip>
---
 dp/knapsack.cc | 28 ++++++++++++++++++++++++++++
 1 file changed, 28 insertions(+)
 create mode 100644 dp/knapsack.cc

diff --git a/dp/knapsack.cc b/dp/knapsack.cc
new file mode 100644
index 0000000..be7c5d0
--- /dev/null
+++ b/dp/knapsack.cc
@@ -0,0 +1,28 @@
+// Part of the code from a Cf problem to demonstrate how to dump the path from a knapsack-dp solution.
+for (int j = 1; j <= m; ++j) {
+    auto [t, p, idx] = events[i][j - 1];
+    for (int k = 0; k <= 100; ++k) {
+        dp[j][k] = dp[j - 1][k];
+    }
+    for (int k = 0; k <= 100; ++k) {
+        int new_percent = min(100, dp[j - 1][k].second + p);  // In this problem we just omit the capacity over 100.
+        if (dp[j][new_percent].first > dp[j - 1][k].first + t) {
+            dp[j][new_percent] = {dp[j - 1][k].first + t, dp[j - 1][k].second + p};
+        }
+    }
+}
+ll tm = dp[m][100].first;
+if (tm + start > a[i]) {
+    cout << -1 << '\n';
+    return;
+} else {
+    int k = dp[m][100].second;
+    for (int j = m; j >= 1; --j) {
+        auto [t, p, idx] = events[i][j - 1];
+        if (dp[j][min(100, k)] < dp[j - 1][min(100, k)]) {  // Find the last step of the construction process.
+            k -= p;  // Prepare to find the next one.
+            res.emplace_back(idx);  // Add the current item as part of the solution.
+        }
+    }
+    start += tm;
+}