# subject

## describe

## Enter Description:

## Output Description:

## Input:

90 4 20 25 30 20 40 50 10 18 40 2 25 30 10 8

## Output:

95 38

# analysis

The 0-1 knapsack problem wants to be abstract, but it is really simple compared with the longest increasing subsequence in front

(I want to record a video and analyze it carefully later ~ don't worry)

# Answer (time O(n*m), space O(n*m))

(drift past ~ defeat niuke.com 3%hhh~)

/* ------------------------------------------------- Author: wry date: 2022/3/5 14:54 Description: 0-1bag ------------------------------------------------- */ #include <bits/stdc++.h> using namespace std; const int MAXN = 1000+10; int price[MAXN][MAXN]; //Total value int weight[MAXN]; int value[MAXN]; int main() { int n,m; //n Indicates the number of items m Indicates the maximum capacity of the backpack while (cin >> m >> n) { memset(price,0, sizeof(price)); for (int i=1;i<=n;i++) { cin >> weight[i] >> value[i]; } for (int i=1;i<=n;i++) { for (int j=1;j<=m;j++) { if (j<weight[i]) { price[i][j] = price[i-1][j]; } else { price[i][j] = max(price[i-1][j-weight[i]]+value[i],price[i-1][j]); //Judge whether it is sacrificing space i Is the total value of the goods high or not i What is the capacity of the backpack j High value of } } } cout << price[n][m] << endl; } return 0; }

# Optimization code (time O(n*m), space O(m))

(the optimization is very ingenious, because in the above code, the price value of each line only depends on the value of the previous line (max(price[i-1][j-weight[i]]+value[i],price[i-1][j]). It can be clearly seen that the I line depends on the i-1 line. Therefore, the optimization scheme is to no longer use the two-dimensional array, only use the one-dimensional array, and update from back to front every time, In this way, there will be no overwrite error! I can only say - wonderful ~ (more than 80%hhhhhh ~)

/* ------------------------------------------------- Author: wry date: 2022/3/5 14:54 Description: 0-1bag ------------------------------------------------- */ #include <bits/stdc++.h> using namespace std; const int MAXN = 1000+10; int price[MAXN]; //Total value(Update from back to front) int weight[MAXN]; int value[MAXN]; int main() { int n,m; //n Indicates the number of items m Indicates the maximum capacity of the backpack while (cin >> m >> n) { memset(price,0, sizeof(price)); for (int i=1;i<=n;i++) { cin >> weight[i] >> value[i]; } for (int i=1;i<=n;i++) { for (int j=m;j>=1;j--) { //Update from back to front every time if (j<weight[i]) { price[j] = price[j]; } else { price[j] = max(price[j-weight[i]]+value[i],price[j]); //Judge whether it is sacrificing space i Is the total value of the goods high or not i What is the capacity of the backpack j High value of } } } cout << price[m] << endl; } return 0; }