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HuangHai
2025-08-30 18:35:01 +08:00
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7
C++专题课程/背包九讲/TestProject/.idea/encodings.xml generated Normal file
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7
C++专题课程/背包九讲/TestProject/.idea/misc.xml generated Normal file
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8
C++专题课程/背包九讲/TestProject/.idea/modules.xml generated Normal file
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131
C++专题课程/背包九讲/TestProject/.idea/workspace.xml generated Normal file
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<project version="4">
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<method v="2">
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<method v="2">
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<method v="2">
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13
C++专题课程/背包九讲/TestProject/1.txt Normal file
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2
4 10
2 4
3 5
4 6
5 10
5 20
3 2
7 3
10 5
15 6
16 10

50
C++专题课程/背包九讲/TestProject/1_01Package.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
#define maxV 1000
/*
1. 01背包
1）题目：
有n件物品和一个容量为v的背包。
第i件物品的费用是c[i]，价值是w[i]。
求解将哪些物品装入背包可使价值总和最大
2）输入：
测试用例数
物品数 背包大小
n个物品的ci和wi
* */
int main(void) {
int times, n, v, ci, wi;
int f[maxV];
//读取数据文件
freopen("../1.txt", "r", stdin);
//读入次数
cin >> times; //本示例是2次
//开始两次循环
while (times--) {
//初始值
memset(f, 0, sizeof(f));
//读取物品个数与背包体积
cin >> n >> v;
//对每个物品进行读取判断
for (int i = 0; i < n; i++) {
//第i件物品的费用(此例出指体积)是c[i]，价值是w[i]。
cin >> ci >> wi;
//利用01背包模板进行填充f数组,动态规划
for (int j = v; j >= 0; j--) { //从大到小
if (j >= ci)
f[j] = max(f[j - ci] + wi, f[j]);
}
}
//输出动态规划的数组
for (int i = 0; i <= v; i++) {
cout << f[i] << " ";
}
cout << endl;
//最后的结果
cout << f[v] << endl;
}
}

13
C++专题课程/背包九讲/TestProject/2.txt Normal file
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2
4 10
2 4
3 5
4 6
5 10
5 20
3 2
7 3
10 5
15 6
16 10

33
C++专题课程/背包九讲/TestProject/2_wholePackage.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
#define maxV 1000
/**
* 2. 完全背包
1）题目：
有n件物品和一个容量为v的背包，每种物品都有无限件可用。
第i种物品的费用是c[i]，价值是w[i]。
求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量，且价值总和最大。
2）输入：
测试用例数
物品数 背包容量
n个物品的ci和vi
*/
int main(void) {
int cases, n, v, ci, wi;
int f[maxV];
freopen("../2.txt", "r", stdin);
cin >> cases;
while (cases--) {
memset(f, 0, sizeof(f));
cin >> n >> v;
for (int i = 0; i < n; i++) {
cin >> ci >> wi;
for (int j = 0; j <= v; j++) {
if (j >= ci)
f[j] = max(f[j], f[j - ci] + wi);
}
}
cout << f[v] << endl;
}
}

11
C++专题课程/背包九讲/TestProject/3.txt Normal file
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2
3 10
1 1 10
2 2 4
3 3 11
4 20
5 2 5
2 3 6
3 4 8
1 6 19

58
C++专题课程/背包九讲/TestProject/3_multiplexPackage.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
#define maxV 1000
int f[maxV], v;
/*
3. 多重背包
1）题目：
有n种物品和一个容量为v的背包。
第i种物品最多有n[i]件可用，每件费用是c[i]，价值是w[i]。
求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量，且价值总和最大。
2）输入：
测试用例数
物品数 背包容量
n个物品的ni ci wi
* */
void ZeroOnePack(int ci, int wi) {
for (int j = v; j >= 0; j--)
if (j >= ci)
f[j] = max(f[j], f[j - ci] + wi);
}
void CompletePack(int ci, int wi) {
for (int j = 0; j <= v; j++)
if (j >= ci)
f[j] = max(f[j], f[j - ci] + wi);
}
void MultiplePack(int ni, int ci, int wi) {
if (ni * ci >= v) {
CompletePack(ci, wi);
return;
}
int k = 1, amount = ni;
while (k < ni) {
ZeroOnePack(ci * k, wi * k);
amount -= k;
k *= 2;
}
ZeroOnePack(ci * amount, wi * amount);
}
int main(void) {
int cases, n, ni, ci, wi;
freopen("../3.txt", "r", stdin);
cin >> cases;
while (cases--) {
memset(f, 0, sizeof(f));
cin >> n >> v;
for (int i = 0; i < n; i++) {
cin >> ni >> ci >> wi;
MultiplePack(ni, ci, wi);
}
//for (int i = 0; i <= v; i++) cout << f[i] << " "; cout << endl;
cout << f[v] << endl;
}
return 0;
}

11
C++专题课程/背包九讲/TestProject/4.txt Normal file
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2
3 10
-1 1 2
2 2 5
1 3 7
4 20
3 2 5
4 4 6
-1 6 8
1 7 10

61
C++专题课程/背包九讲/TestProject/4_mixPackage.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
/**
* 4. 混合三种背包
1）问题：
有的物品只可以取一次（01背包），有的物品可以取无限次（完全背包），有的物品可以取的次数有一个上限（多重背包）。
应该怎么求解呢？
2）输入：
测试用例数
物品数 背包容量
第i种物品的ni(无限次的标为-1) ci wi
*/
#define maxV 1000
int f[maxV], v;
void ZeroOnePack(int ci, int wi) {
for (int j = v; j >= 0; j--)
if (j >= ci)
f[j] = max(f[j], f[j - ci] + wi);
}
void CompletePack(int ci, int wi) {
for (int j = 0; j <= v; j++)
if (j >= ci)
f[j] = max(f[j], f[j - ci] + wi);
}
void MultiplePack(int ni, int ci, int wi) {
if (ni * ci >= v) {
CompletePack(ci, wi);
return;
}
int k = 1, amount = ni;
while (k < ni) {
ZeroOnePack(ci * k, wi * k);
amount -= k;
k *= 2;
}
ZeroOnePack(ci * amount, wi * amount);
}
int main(void) {
int cases, n, ni, ci, wi;
freopen("../4.txt", "r", stdin);
cin >> cases;
while (cases--) {
memset(f, 0, sizeof(f));
cin >> n >> v;
for (int i = 0; i < n; i++) {
cin >> ni >> ci >> wi;
if (ni == 1) ZeroOnePack(ci, wi);
else if (ni == -1) CompletePack(ci, wi);
else MultiplePack(ni, ci, wi);
}
//for (int i = 0; i <= v; i++) cout << f[i] << " "; cout << endl;
cout << f[v] << endl;
}
return 0;
}

11
C++专题课程/背包九讲/TestProject/5.txt Normal file
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2
3 10 5
1 2 4
2 1 4
4 4 6
4 5 6
1 1 3
1 2 4
2 1 4
3 4 7

44
C++专题课程/背包九讲/TestProject/5_TwoDimensionPackage.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
/**
* 5. <20><>ά<EFBFBD><CEAC><EFBFBD>õı<C3B5><C4B1><EFBFBD>
1<><31><EFBFBD><EFBFBD><EFBFBD>⣺
<20><><EFBFBD><EFBFBD>ÿ<EFBFBD><C3BF><EFBFBD><EFBFBD>Ʒ<EFBFBD><C6B7><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ֲ<EFBFBD>ͬ<EFBFBD>ķ<EFBFBD><C4B7>ã<EFBFBD>
ѡ<><D1A1><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>Ʒ<EFBFBD><C6B7><EFBFBD><EFBFBD>ͬʱ<CDAC><CAB1><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ִ<EFBFBD><D6B4>ۣ<EFBFBD>
<20><><EFBFBD><EFBFBD>ÿ<EFBFBD>ִ<EFBFBD><D6B4>۶<EFBFBD><DBB6><EFBFBD>һ<EFBFBD><D2BB><EFBFBD>ɸ<EFBFBD><C9B8><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ֵ<EFBFBD><D6B5><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>
<20><><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ѡ<EFBFBD><D1A1><EFBFBD><EFBFBD>Ʒ<EFBFBD><C6B7><EFBFBD>Եõ<D4B5><C3B5><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ļ<EFBFBD>ֵ<EFBFBD><D6B5>
<20><>i<EFBFBD><69><EFBFBD><EFBFBD>Ʒ<EFBFBD><C6B7><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ִ<EFBFBD><D6B4>۷ֱ<DBB7>Ϊa[i]<5D><>b[i]<5D><>
<20><><EFBFBD>ִ<EFBFBD><D6B4>ۿɸ<DBBF><C9B8><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ֵ<EFBFBD><D6B5><EFBFBD><EFBFBD><EFBFBD>ֱ<EFBFBD><D6B1><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ֱ<EFBFBD>ΪV<CEAA><56>U<EFBFBD><55><EFBFBD><EFBFBD>Ʒ<EFBFBD>ļ<EFBFBD>ֵΪw[i]<5D><>
2<><32><EFBFBD><EFBFBD><EFBFBD>룺
<20><><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>
<20><>Ʒ<EFBFBD><C6B7> <20><>һ<EFBFBD><D2BB><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>v <20>ڶ<EFBFBD><DAB6><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>u
<20><>i<EFBFBD><69><EFBFBD><EFBFBD>Ʒ<EFBFBD><C6B7>ai bi wi
*/
#define maxV 1000
#define maxU 1000
int main(void) {
int cases, n, v, u, ai, bi, wi;
int f[maxV][maxU];
freopen("../5.txt", "r", stdin);
cin >> cases;
while (cases--) {
memset(f, 0, sizeof(f));
cin >> n >> v >> u;
for (int i = 0; i < n; i++) {
cin >> ai >> bi >> wi;
for (int j = v; j >= 0; j--) {
for (int k = u; k >= 0; k--) {
if (ai <= j && bi <= k)
f[j][k] = max(f[j][k], f[j - ai][k - bi] + wi);
}
}
}
// for (int i = 0; i <= v; i++)
// for(int j=0;j<=u;j++)
// cout << f[i][j] << " ";
// cout << endl;
cout << f[v][u] << endl;
}
}

13
C++专题课程/背包九讲/TestProject/6.txt Normal file
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@@ -0,0 +1,13 @@
2
4 10 2
3 5 0
4 6 0
2 7 1
1 6 1
5 10 3
1 3 0
2 4 0
3 5 1
4 7 1
5 8 2

46
C++专题课程/背包九讲/TestProject/6_GroupPackage.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
/**
6. 分组背包
1）问题：
有n件物品和一个容量为v的背包。
第i件物品的费用是c[i]，价值是w[i]。
这些物品被划分为若干组，每组中的物品互相冲突，最多选一件。
求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量，且价值总和最大。
2）输入：
测试用例数
物品数 背包容量 组数g(组号范围：0 ~ g-1)
第i件物品的ci wi gi(所属组号)
*/
using namespace std;
#define maxV 1000
#define maxG 100
#define maxN 100
int main(void) {
int cases, n, v, g, gi;
int f[maxV];
int ci[maxN], wi[maxN];
freopen("../6.txt", "r", stdin);
cin >> cases;
while (cases--) {
memset(f, 0, sizeof(f));
cin >> n >> v >> g;
vector<vector<int> > gMap(g);
for (int i = 0; i < n; i++) {
cin >> ci[i] >> wi[i] >> gi;
gMap[gi].push_back(i);
}
for (int i = 0; i < g; i++) {
for (int j = v; j >= 0; j--) {
for (int k = 0; k < gMap[i].size(); k++) {
if (j >= ci[gMap[i][k]])
f[j] = max(f[j], f[j - ci[gMap[i][k]]] + wi[gMap[i][k]]);
}
}
}
cout << f[v] << endl;
}
}

11
C++专题课程/背包九讲/TestProject/7.txt Normal file
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@@ -0,0 +1,11 @@
2
3 10
3 5 -1
4 6 0
7 10 0
4 15
4 1 -1
6 8 0
7 10 0
10 4 -1

84
C++专题课程/背包九讲/TestProject/7_DependOnPackage.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
/**
7. 有依赖的背包
1）题目：
这种背包问题的物品间存在某种“依赖”的关系。
也就是说，i依赖于j，表示若选物品i，则必须选物品j。
为了简化起见，我们先设没有某个物品既依赖于别的物品，又被别的物品所依赖；
另外，没有某件物品同时依赖多件物品。
2）输入：
测试用例数
物品数 背包大小
第i个物品的ci wi di(依赖物品的编号，-1为不依赖其他物品)
*/
#define maxV 1000
#define maxG 100
int main(void) {
int cases, n, v, ci[maxV], wi[maxV], di, f[maxV];
freopen("../7.txt", "r", stdin);
cin >> cases;
while (cases--) {
memset(f, 0, sizeof(f));
cin >> n >> v;
// group[i]空表示i号物品有依赖，因此存放到其他组里
// 只有一个元素表示i号物品既无依赖也不被依赖
// 有多个元素表示i号物品被依赖，这里自己的编号i也被存放进group[i]
vector<vector<int> > groups(n);
// 读入数据并储存起来
for (int i = 0; i < n; i++) {
cin >> ci[i] >> wi[i] >> di;
if (di == -1) groups[i].push_back(i);
else groups[di].push_back(i);
}
// 对每个有多个元素的组进行01背包
for (int i = 0; i < n; i++) {
if (groups[i].size() > 1) {
vector<int> giOri; //group[i]的拷贝，排除i本身
int newV = v - ci[i];
// 复制group[i]中的元素，排除i
for (int j = 0; j < groups[i].size(); j++) {
if (groups[i][j] != i)
giOri.push_back(groups[i][j]);
}
// 把等效物品组存入group[i]中
groups[i].clear();
groups[i].resize(newV + 1, 0);
for (int j = 0; j < giOri.size(); j++) {
for (int k = newV; k >= 0; k--) {
if (ci[giOri[j]] <= k) {
groups[i][k] = max(groups[i][k], groups[i][k - ci[giOri[j]]] + wi[giOri[j]]);
}
}
}
}
}
// 进行分组背包
for (int i = 0; i < n; i++) {
if (groups[i].size() == 0) continue;
else if (groups[i].size() == 1) { // i物品无依赖且不被依赖
for (int j = v; j >= 0; j--) {
if (j >= ci[i])
f[j] = max(f[j], f[j - ci[i]] + wi[i]);
}
} else { // i物品被依赖, i组第k个物品的cost为k + ci[i], weight为group[i][k] + wi[i]
for (int j = v; j >= 0; j--) {
for (int k = 0; k < groups[i].size(); k++) {
if (j >= k + ci[i])
f[j] = max(f[j], f[j - k - ci[i]] + groups[i][k] + wi[i]);
}
}
}
}
cout << f[v] << endl;
}
}

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C++专题课程/背包九讲/TestProject/8.txt Normal file
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2
5 5 0
1 2 -1
2 7 0
2 5 0
2 6 2
1 4 2
6 5 0
1 2 -1
1 3 0
2 3 0
2 4 0
2 7 2
3 10 2

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C++专题课程/背包九讲/TestProject/8_GeneralizationPackage.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
/**
8. 泛化物品
）题目：
在背包容量为v的背包问题中，泛化物品是一个定义域为0..v中的整数的函数h，当分配给它的费用为v时，能得到的价值就是h(v)。
为了应用泛化物品思想，这里假设题目为：存在依赖关系树（简单起见，只有一个根节点），即有依赖的物品也可以被依赖，可结合原文7.3节来理解此句。
2）输入：
测试用例数
物品数 背包容量 根节点编号
第i个物品的ci wi di(依赖物品的编号，-1为不依赖其他物品)
* @return
*/
#define maxN 100
#define maxV 1000
int n, v;
int cnt = 0;
int head[maxN];
int wi[maxN], ci[maxN];
int f[maxN][maxV];
struct Edge {
int v, next;
} e[maxN - 1];
void addEdge(int u, int v) {
e[cnt].v = v;
e[cnt].next = head[u];
head[u] = cnt++;
}
void treeDP(int u) {
for (int i = ci[u]; i <= v; i++) {
f[u][i] = wi[u];
}
for (int i = head[u]; i != -1; i = e[i].next) {
int curr = e[i].v;
treeDP(curr);
for (int j = v; j >= 0; j--) {
for (int k = j - ci[u]; k >= 0; k--) {
f[u][j] = max(f[u][j], f[u][j - k] + f[curr][k]);
}
}
}
}
int main(void) {
int cases, root;
freopen("../8.txt", "r", stdin);
cin >> cases;
while (cases--) {
cnt = 0;
memset(head, -1, sizeof(head));
memset(f, 0, sizeof(f));
cin >> n >> v >> root;
for (int i = 0; i < n; i++) {
int di;
cin >> ci[i] >> wi[i] >> di;
addEdge(di, i);
}
treeDP(root);
cout << f[root][v] << endl;
}
}

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C++专题课程/背包九讲/TestProject/9.txt Normal file
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2
4 10
2 4
3 5
4 6
5 10
5 20
3 2
7 3
10 5
15 6
16 10

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C++专题课程/背包九讲/TestProject/9_ChangePackage.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
/**
*9. 背包问题的变化
1）题目：
输出01背包的具体方案
2）输入：
同01背包
* @return
*/
#define maxV 1000
#define maxN 100
int main(void) {
int cases, n, v, ci[maxN], wi;
int f[maxV];
bool g[maxN][maxV]; //g[i][v]=0 表示没放i时的f(i, v)较大，
//g[i][v]=1 表示放进i时的f(i, v)较大
freopen("../9.txt", "r", stdin);
cin >> cases;
while (cases--) {
memset(f, 0, sizeof(f));
memset(g, 0, sizeof(g));
cin >> n >> v;
for (int i = 0; i < n; i++) {
cin >> ci[i] >> wi;
for (int j = v; j >= 0; j--) {
if (j >= ci[i]) {
if (f[j - ci[i]] + wi > f[j]) {
f[j] = f[j - ci[i]] + wi;
g[i][j] = 1;
}
}
}
}
int i = n - 1, j = v;
while (i >= 0) {
if (g[i][j] == 1) {
cout << "选了" << i << endl;
j -= ci[i];
} else {
cout << "没选" << i << endl;
}
i--;
}
cout << endl;
}
}

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C++专题课程/背包九讲/TestProject/CMakeLists.txt Normal file
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cmake_minimum_required(VERSION 3.15)
project(TestProject)
set(CMAKE_CXX_STANDARD 11)
set(CMAKE_CXX FLAGS -fexec-charset=GBK)
#set(CMAKE_CXX_FLAGS "${CMAKE_CXX_FLAGS} -std=c++11")
# 修改clion 内存
MATH(EXPR stack_size "4*1024*1024")
set(CMAKE_EXE_LINKER_FLAGS "-Wl,--stack,${stack_size}")
add_executable(TestProject 1_01Package.cpp)
add_executable(2_wholePackage 2_wholePackage.cpp)
add_executable(3_multiplexPackage 3_multiplexPackage.cpp)
add_executable(4_mixPackage 4_mixPackage.cpp)
add_executable(5_TwoDimensionPackage 5_TwoDimensionPackage.cpp)
add_executable(6_GroupPackage 6_GroupPackage.cpp)
add_executable(7_DependOnPackage 7_DependOnPackage.cpp)
add_executable(8_GeneralizationPackage 8_GeneralizationPackage.cpp)
add_executable(9_ChangePackage 9_ChangePackage.cpp)

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C++专题课程/背包九讲/TestProject/背包九讲C++实现完整代码.txt Normal file
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https://blog.csdn.net/desirepath/article/details/71679170

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C++专题课程/背包九讲/背包问题九讲.pdf Normal file
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C++专题课程/背包九讲/阅读资料.txt Normal file
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https://blog.csdn.net/desirepath/article/details/71679170
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https://www.cnblogs.com/BrokenString/p/9279544.html