贪心解决背包问题c++语言
时间: 2023-09-16 11:13:27 浏览: 51
好的,我可以回答关于贪心解决背包问题的问题。背包问题是一个经典的组合优化问题,贪心算法可以用于求解该问题。在贪心算法中,我们优先选择价值最高的物品放入背包,直到背包装满或没有物品可选。这种贪心策略可以得到较优的结果,但并不一定是最优解。在实际应用中,还需要结合动态规划等算法来解决更为复杂的背包问题。
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贪心法背包问题c++
以下是使用贪心算法解决背包问题的C++代码示例:
```cpp
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct Item {
int weight;
int value;
double ratio;
};
bool compare(Item a, Item b) {
return a.ratio > b.ratio;
}
double knapsack(int capacity, vector<Item>& items) {
sort(items.begin(), items.end(), compare);
double totalValue = 0.0;
int currentWeight = 0;
for (int i = 0; i < items.size(); i++) {
if (currentWeight + items[i].weight <= capacity) {
currentWeight += items[i].weight;
totalValue += items[i].value;
} else {
int remainingCapacity = capacity - currentWeight;
totalValue += items[i].ratio * remainingCapacity;
break;
}
}
return totalValue;
}
int main() {
int capacity = 10;
vector<Item> items = {{2, 10}, {3, 5}, {5, 15}, {7, 7}, {1, 6}, {4, 18}, {1, 3}};
double maxValue = knapsack(capacity, items);
cout << "Maximum value that can be carried: " << maxValue << endl;
return 0;
}
```
贪心法背包问题C++
以下是使用贪心法解决背包问题的C++代码示例:
```cpp
#include <iostream>
#include <algorithm>
using namespace std;
void Sort(int n, float v[], float w[]) {
// 根据单位重量内物品价值进行排序
float ratio[n];
for (int i = 0; i < n; i++) {
ratio[i] = v[i] / w[i];
}
for (int i = 0; i < n-1; i++) {
for (int j = 0; j < n-i-1; j++) {
if (ratio[j] < ratio[j+1]) {
swap(ratio[j], ratio[j+1]);
swap(v[j], v[j+1]);
swap(w[j], w[j+1]);
}
}
}
}
void Knapsack(int n, float M, float v[], float w[], float x[]) {
Sort(n, v, w);
for (int i = 0; i < n; i++) {
x[i] = 0;
}
float c = M;
for (int i = 0; i < n; i++) {
if (w[i] > c) {
break;
}
x[i] = 1;
c -= w[i];
}
if (i < n) {
x[i] = c / w[i];
}
}
int main() {
int n = 5; // 物品数
float M = 10; // 背包大小
float v[] = {60, 100, 120, 80, 70}; // 物品价值数组
float w[] = {10, 20, 30, 40, 50}; // 物品重量数组
float x[n]; // 占比数组
Knapsack(n, M, v, w, x);
cout << "物品占比:";
for (int i = 0; i < n; i++) {
cout << x[i] << " ";
}
cout << endl;
return 0;
}
```