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* Retranslate the complete English edition from Simplified Chinese * Polish three English explanations after reader-focused review
110 lines
3.7 KiB
C++
110 lines
3.7 KiB
C++
#include <algorithm>
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#include <iostream>
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#include <vector>
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using namespace std;
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/* 0-1 knapsack: brute-force search */
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int knapsackDFS(vector<int> &wgt, vector<int> &val, int i, int c) {
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// If all items have been considered or no capacity remains, return value 0
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if (i == 0 || c == 0) {
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return 0;
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}
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// If the item exceeds the remaining capacity, it must be excluded
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if (wgt[i - 1] > c) {
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return knapsackDFS(wgt, val, i - 1, c);
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}
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// Calculate the maximum values obtained by excluding and including item i
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int no = knapsackDFS(wgt, val, i - 1, c);
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int yes = knapsackDFS(wgt, val, i - 1, c - wgt[i - 1]) + val[i - 1];
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// Return the larger value from the two choices
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return max(no, yes);
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}
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/* 0-1 knapsack: memoized search */
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int knapsackDFSMem(vector<int> &wgt, vector<int> &val, vector<vector<int>> &mem, int i, int c) {
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// If all items have been considered or no capacity remains, return value 0
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if (i == 0 || c == 0) {
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return 0;
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}
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// If a result has already been recorded, return it directly
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if (mem[i][c] != -1) {
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return mem[i][c];
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}
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// If the item exceeds the remaining capacity, it must be excluded
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if (wgt[i - 1] > c) {
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return knapsackDFSMem(wgt, val, mem, i - 1, c);
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}
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// Calculate the maximum values obtained by excluding and including item i
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int no = knapsackDFSMem(wgt, val, mem, i - 1, c);
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int yes = knapsackDFSMem(wgt, val, mem, i - 1, c - wgt[i - 1]) + val[i - 1];
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// Record and return the larger value from the two choices
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mem[i][c] = max(no, yes);
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return mem[i][c];
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}
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/* 0-1 knapsack: dynamic programming */
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int knapsackDP(vector<int> &wgt, vector<int> &val, int cap) {
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int n = wgt.size();
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// Initialize the dp table
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vector<vector<int>> dp(n + 1, vector<int>(cap + 1, 0));
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// State transitions
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for (int i = 1; i <= n; i++) {
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for (int c = 1; c <= cap; c++) {
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if (wgt[i - 1] > c) {
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// If item i exceeds the remaining capacity, do not choose it
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dp[i][c] = dp[i - 1][c];
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} else {
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// Take the larger value obtained by excluding and including item i
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dp[i][c] = max(dp[i - 1][c], dp[i - 1][c - wgt[i - 1]] + val[i - 1]);
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}
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}
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}
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return dp[n][cap];
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}
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/* 0-1 knapsack: space-optimized dynamic programming */
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int knapsackDPComp(vector<int> &wgt, vector<int> &val, int cap) {
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int n = wgt.size();
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// Initialize the dp table
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vector<int> dp(cap + 1, 0);
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// State transitions
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for (int i = 1; i <= n; i++) {
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// Traverse in decreasing order
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for (int c = cap; c >= 1; c--) {
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if (wgt[i - 1] <= c) {
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// Take the larger value obtained by excluding and including item i
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dp[c] = max(dp[c], dp[c - wgt[i - 1]] + val[i - 1]);
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}
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}
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}
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return dp[cap];
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}
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/* Driver code */
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int main() {
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vector<int> wgt = {10, 20, 30, 40, 50};
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vector<int> val = {50, 120, 150, 210, 240};
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int cap = 50;
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int n = wgt.size();
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// Brute-force search
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int res = knapsackDFS(wgt, val, n, cap);
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cout << "Maximum total item value within the knapsack capacity: " << res << endl;
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// Memoized search
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vector<vector<int>> mem(n + 1, vector<int>(cap + 1, -1));
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res = knapsackDFSMem(wgt, val, mem, n, cap);
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cout << "Maximum total item value within the knapsack capacity: " << res << endl;
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// Dynamic programming
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res = knapsackDP(wgt, val, cap);
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cout << "Maximum total item value within the knapsack capacity: " << res << endl;
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// Space-optimized dynamic programming
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res = knapsackDPComp(wgt, val, cap);
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cout << "Maximum total item value within the knapsack capacity: " << res << endl;
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return 0;
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}
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