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pull merged 1 commit into AlgorithmAndLeetCode:master from youngyangyang04:master
Oct 4, 2023
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1 change: 1 addition & 0 deletions README.md
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Expand Up @@ -74,6 +74,7 @@

* 编程语言
* [C++面试&C++学习指南知识点整理](https://github.com/youngyangyang04/TechCPP)
* [C++语言基础课](https://kamacoder.com/course.php?course_id=1)

* 项目
* [基于跳表的轻量级KV存储引擎](https://github.com/youngyangyang04/Skiplist-CPP)
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2 changes: 1 addition & 1 deletion problems/周总结/20201107回溯周末总结.md
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Expand Up @@ -31,7 +31,7 @@ for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target;

在[回溯算法:求组合总和(二)](https://programmercarl.com/0039.组合总和.html)第一个树形结构没有画出startIndex的作用,**这里这里纠正一下,准确的树形结构如图所示:**

![39.组合总和](https://code-thinking-1253855093.file.myqcloud.com/pics/20201123202227835.png)
![39.组合总和](https://code-thinking-1253855093.file.myqcloud.com/pics/20201223170730367.png)

## 周二

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2 changes: 1 addition & 1 deletion problems/回溯总结.md
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Expand Up @@ -116,8 +116,8 @@ void backtracking(参数) {
**注意以上我只是说求组合的情况,如果是排列问题,又是另一套分析的套路**。

树形结构如下:
![39.组合总和](https://code-thinking-1253855093.file.myqcloud.com/pics/20201223170730367.png)

![39.组合总和](https://code-thinking-1253855093.file.myqcloud.com/pics/20201118152521990.png)

最后还给出了本题的剪枝优化,如下:

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52 changes: 52 additions & 0 deletions problems/背包理论基础01背包-1.md
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Expand Up @@ -8,6 +8,9 @@

# 动态规划:01背包理论基础


本题力扣上没有原题,大家可以去[卡码网第46题](https://kamacoder.com/problem.php?id=1046)去练习,题意是一样的。

## 算法公开课

**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[带你学透0-1背包问题!](https://www.bilibili.com/video/BV1cg411g7Y6/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
Expand Down Expand Up @@ -261,6 +264,55 @@ int main() {

```

本题力扣上没有原题,大家可以去[卡码网第46题](https://kamacoder.com/problem.php?id=1046)去练习,题意是一样的,代码如下:

```CPP

//二维dp数组实现
#include <bits/stdc++.h>
using namespace std;

int n, bagweight;// bagweight代表行李箱空间
void solve() {
vector<int> weight(n, 0); // 存储每件物品所占空间
vector<int> value(n, 0); // 存储每件物品价值
for(int i = 0; i < n; ++i) {
cin >> weight[i];
}
for(int j = 0; j < n; ++j) {
cin >> value[j];
}
// dp数组, dp[i][j]代表行李箱空间为j的情况下,从下标为[0, i]的物品里面任意取,能达到的最大价值
vector<vector<int>> dp(weight.size(), vector<int>(bagweight + 1, 0));

// 初始化, 因为需要用到dp[i - 1]的值
// j < weight[0]已在上方被初始化为0
// j >= weight[0]的值就初始化为value[0]
for (int j = weight[0]; j <= bagweight; j++) {
dp[0][j] = value[0];
}

for(int i = 1; i < weight.size(); i++) { // 遍历科研物品
for(int j = 0; j <= bagweight; j++) { // 遍历行李箱容量
// 如果装不下这个物品,那么就继承dp[i - 1][j]的值
if (j < weight[i]) dp[i][j] = dp[i - 1][j];
// 如果能装下,就将值更新为 不装这个物品的最大值 和 装这个物品的最大值 中的 最大值
// 装这个物品的最大值由容量为j - weight[i]的包任意放入序号为[0, i - 1]的最大值 + 该物品的价值构成
else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
}
}
cout << dp[weight.size() - 1][bagweight] << endl;
}

int main() {
while(cin >> n >> bagweight) {
solve();
}
return 0;
}

```


## 总结

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47 changes: 46 additions & 1 deletion problems/背包理论基础01背包-2.md
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Expand Up @@ -6,14 +6,14 @@

# 动态规划:01背包理论基础(滚动数组)

本题力扣上没有原题,大家可以去[卡码网第46题](https://kamacoder.com/problem.php?id=1046)去练习

## 算法公开课

**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[带你学透0-1背包问题!(滚动数组)](https://www.bilibili.com/video/BV1BU4y177kY/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。

## 思路


昨天[动态规划:关于01背包问题,你该了解这些!](https://programmercarl.com/背包理论基础01背包-1.html)中是用二维dp数组来讲解01背包。

今天我们就来说一说滚动数组,其实在前面的题目中我们已经用到过滚动数组了,就是把二维dp降为一维dp,一些录友当时还表示比较困惑。
Expand Down Expand Up @@ -159,6 +159,8 @@ dp[1] = dp[1 - weight[0]] + value[0] = 15



C++代码如下:

```CPP
void test_1_wei_bag_problem() {
vector<int> weight = {1, 3, 4};
Expand All @@ -181,6 +183,49 @@ int main() {

```

本题力扣上没有原题,大家可以去[卡码网第46题](https://kamacoder.com/problem.php?id=1046)去练习,题意是一样的,代码如下:

```CPP
// 一维dp数组实现
#include <iostream>
#include <vector>
using namespace std;

int main() {
// 读取 M 和 N
int M, N;
cin >> M >> N;

vector<int> costs(M);
vector<int> values(M);

for (int i = 0; i < M; i++) {
cin >> costs[i];
}
for (int j = 0; j < M; j++) {
cin >> values[j];
}

// 创建一个动态规划数组dp,初始值为0
vector<int> dp(N + 1, 0);

// 外层循环遍历每个类型的研究材料
for (int i = 0; i < M; ++i) {
// 内层循环从 N 空间逐渐减少到当前研究材料所占空间
for (int j = N; j >= costs[i]; --j) {
// 考虑当前研究材料选择和不选择的情况,选择最大值
dp[j] = max(dp[j], dp[j - costs[i]] + values[i]);
}
}

// 输出dp[N],即在给定 N 行李空间可以携带的研究材料最大价值
cout << dp[N] << endl;

return 0;
}

```

可以看出,一维dp 的01背包,要比二维简洁的多! 初始化 和 遍历顺序相对简单了。

**所以我倾向于使用一维dp数组的写法,比较直观简洁,而且空间复杂度还降了一个数量级!**
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39 changes: 39 additions & 0 deletions problems/背包问题理论基础完全背包.md
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Expand Up @@ -7,6 +7,8 @@

# 动态规划:完全背包理论基础

本题力扣上没有原题,大家可以去[卡码网第52题](https://kamacoder.com/problem.php?id=1046)去练习,题意是一样的。

## 算法公开课

**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[带你学透完全背包问题! ](https://www.bilibili.com/video/BV1uK411o7c9/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
Expand Down Expand Up @@ -162,6 +164,43 @@ int main() {

```

本题力扣上没有原题,大家可以去[卡码网第46题](https://kamacoder.com/problem.php?id=1046)去练习,题意是一样的,C++代码如下:

```cpp
#include <iostream>
#include <vector>
using namespace std;

// 先遍历背包,再遍历物品
void test_CompletePack(vector<int> weight, vector<int> value, int bagWeight) {

vector<int> dp(bagWeight + 1, 0);

for(int j = 0; j <= bagWeight; j++) { // 遍历背包容量
for(int i = 0; i < weight.size(); i++) { // 遍历物品
if (j - weight[i] >= 0) dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
}
}
cout << dp[bagWeight] << endl;
}
int main() {
int N, V;
cin >> N >> V;
vector<int> weight;
vector<int> value;
for (int i = 0; i < N; i++) {
int w;
int v;
cin >> w >> v;
weight.push_back(w);
value.push_back(v);
}
test_CompletePack(weight, value, V);
return 0;
}
```



## 总结

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AltStyle によって変換されたページ (->オリジナル) /