Knapsack
01 Knapsack
pull dp
dp[i, j] = maximum total value with total cost j using items 0..=i
dp[i, j] = max(
dp[i - 1, j], // don't use item
dp[i - 1, j - w[i]] + v[i], // use item i
)
Unbounded Knapsack
dp[i, j] = maximum total value of total weight j using items 0..=i
if we use item i one time, dp[i, j] = dp[i, j - w[i]] + v[i]
if we don't use item i, dp[i, j] = dp[i - 1, j]
Easier to code if reusing the dp:
let mut dp = vec![inf; w + 1];
dp[0] = 0;
for i in 0..vs.len() {
for j in vs[i]..=p {
dp[j] = dp[j].min(dp[j - vs[i]] + ws[i]);
}
}Bounded Knapsack
With deque optimization.
Large Weight with Small Value
dp[i, j] = minimum total weight of total value j using items 0..=i
if we use item i one time, dp[i, j] = dp[i, j - v[i]] + c[i]
if we don't use item i, dp[i, j] = dp[i - 1, j]
answer is (0..=max_v).filer(|&v| dp[n - 1][v] <= w).max().unwrap().
AtCoder DP E AtCoder Sumitrust 2019 C ABC153E
Upperbound Clipping
Given N positive numbers, can we pick some of them such that the total is at least
K?
Easier to code if using push-dp instead of pull-dp.
dp[j + w] <- dp[j] + v
dp[(j + w).clip(x)] <- dp[j] + v
Subset Sum
dp[i, j] = is it possible to make total j using items 0..=i
dp[i, j] = dp[i - 1, j] or dp[i - 1, j - arr[i]]