动态规划（二）0-1背包问题-优快云博客

这篇博客探讨了如何使用动态规划方法解决经典的0-1背包问题。作者提供了四种不同的实现方式，包括递归记忆化搜索和动态规划表格填充，详细分析了它们的时间复杂度和空间复杂度。每种方法都通过实例展示了如何找到在给定容量限制下，能装入物品的最大价值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

0-1背包问题

递归+记忆中间结果

#include <iostream>
#include "string"
#include<vector>
using namespace std;
int backage01(vector<int> w, vector<int> v, int contains, int index);
vector<vector<int>> memory;
int backage01(vector<int> w, vector<int> v, int contains) {
    assert(w.size() == v.size());
    if (w.size() == 0) return 0;
    memory = vector<vector<int>> (v.size(), vector<int> (contains+1, -1));
    return backage01(w, v, contains, w.size()-1);
}


//状态：背包容量为index, 装入的价值最高
//状态转移 f(i, C) = f(i-1, C)
//                  f(i-1, C-w[i]) + v[i]
int backage01(vector<int> w, vector<int> v, int contains, int index) {
    if (index < 0 || contains < 0) return 0;
    if (memory[index][contains] != -1) return memory[index-1][contains];
    int res = backage01(w, v, contains, index-1);
    if (contains - w[index] >= 0) {
        res = max(res, backage01(w, v, contains-w[index], index-1) + v[index]);
    }
    memory[index][contains] = res;
    return res;
}

int main() {
    int weights[] = {1, 2, 3};
    vector<int> w(weights, weights + 3);
    int values[] = {6, 10, 12};
    vector<int> v(values, values + 3);
    int contains = 5;
    int res = backage01(w, v, contains);
    cout << res << endl;
}

动态规划（一）

时间复杂度（O(n * C)）
空间复杂度（O(n * C)）

#include <iostream>
#include "string"
#include<vector>
using namespace std;
//状态：背包容量为index, 装入的价值最高
//状态转移 f(i, C) = f(i-1, C)
//                  f(i-1, C-w[i]) + v[i]

int backage01(vector<int> w, vector<int> v, int contains) {
    assert(w.size() == v.size());
    if (w.size() == 0) return 0;
    int n = w.size();
    vector<vector<int>> memory(n, vector<int>(contains+1, 0));
    for (int i = 0; i <= contains; i++) {
        if (i >= w[0]) memory[0][i] = v[0];
        else memory[0][i] = 0;
    }
    //打印中间结果
    for (int i = 1; i < n; i++) {
        for (int j = 1; j <= contains; j++) {
            int res = memory[i-1][j];
            if (j >= w[i]) {
                res = max(res, memory[i-1][j-w[i]] + v[i]);
            }
            memory[i][j] = res;
        }
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= contains; j++) {
            cout << memory[i][j] << " ";
        }
        cout << endl;
    }
    return memory[n-1][contains];
}

int main() {
    int weights[] = {1, 2, 3};
    vector<int> w(weights, weights + 3);
    int values[] = {6, 10, 12};
    vector<int> v(values, values + 3);
    int contains = 5;
    int res = backage01(w, v, contains);
    cout << res << endl;
}

动态规划（二）

时间复杂度（O(n * C)）
空间复杂度（O(2 * C)）

#include <iostream>
#include "string"
#include<vector>
using namespace std;
//状态：背包容量为index, 装入的价值最高
//状态转移 f(i, C) = f(i-1, C)
//                  f(i-1, C-w[i]) + v[i]

int backage01(vector<int> w, vector<int> v, int contains) {
    assert(w.size() == v.size());
    if (w.size() == 0) return 0;
    int n = w.size();
    vector<vector<int>> memory(2, vector<int>(contains+1, 0));
    for (int i = 0; i <= contains; i++) {
        if (i >= w[0]) memory[0][i] = v[0];
        else memory[0][i] = 0;
    }
    
    for (int i = 1; i < n; i++) {
        for (int j = 1; j <= contains; j++) {
            int res = memory[(i-1)%2][j];
            if (j >= w[i]) {
                res = max(res, memory[(i-1)%2][j-w[i]] + v[i]);
            }
            memory[i%2][j] = res;
        }
    }
	//打印中间结果
    for (int i = 0; i < 2; i++) {
        for (int j = 0; j <= contains; j++) {
            cout << memory[i][j] << " ";
        }
        cout << endl;
    }
    return memory[(n-1)%2][contains];
}

int main() {
    int weights[] = {1, 2, 3};
    vector<int> w(weights, weights + 3);
    int values[] = {6, 10, 17};
    vector<int> v(values, values + 3);
    int contains = 5;
    int res = backage01(w, v, contains);
    cout << res << endl;
}

动态规划（二）

时间复杂度（O(n * C)）
空间复杂度（O( C)）

#include <iostream>
#include "string"
#include<vector>
using namespace std;
//状态：背包容量为index, 装入的价值最高
//状态转移 f(i, C) = f(i-1, C)
//                  f(i-1, C-w[i]) + v[i]

int backage01(vector<int> w, vector<int> v, int contains) {
    assert(w.size() == v.size());
    if (w.size() == 0) return 0;
    int n = w.size();
    vector<int> memory(contains+1, 0);
    for (int i = 0; i <= contains; i++) {
        if (i >= w[0]) memory[i] = v[0];
        else memory[i] = 0;
    }

    for (int i = 1; i < n; i++) {
        for (int j = contains; j >= w[i]; j--) {
            int res = memory[j];
            if (j >= w[i]) {
                res = max(res, memory[j-w[i]] + v[i]);
            }
            memory[j] = res;
        }
    }

    //打印中间结果
    for (int j = 0; j <= contains; j++) {
        cout << memory[j] << " ";
    }
    cout << endl;
    return memory[contains];
}

int main() {
    int weights[] = {1, 2, 3};
    vector<int> w(weights, weights + 3);
    int values[] = {6, 10, 17};
    vector<int> v(values, values + 3);
    int contains = 5;
    int res = backage01(w, v, contains);
    cout << res << endl;
}