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.
|
||
|
|
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;
|
||
|
|
}
|