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