湘潭大学算法设计与分析实验回溯动态规划贪心模拟退火解决背包问题

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https://download.csdn.net/download/SQ_ZengYX/88620871

搜索与回溯

#include <iostream>
using namespace std;

#define N 100
int n;
double limitW;
double maxv;
int option[N];
int cop[N];

struct {
	double weight;
	double value;
} a[N];

void BackTrack(int i, double tw, double tv) {
	// 不能超出重量限制
	if (tw > limitW + 0.1)
		return;
	if (tv > maxv) {
		for (int k = 0; k < n; ++k)
			option[k] = cop[k];
		maxv = tv;
	}
	// 选第i个物品
	cop[i] = 1;
	if (i < n)
		BackTrack(i + 1, tw + a[i].weight, tv + a[i].value);
	// 不选第i个物品
	cop[i] = 0;
	if (i < n)
		BackTrack(i + 1, tw, tv);
}


int main() {
	int k;
	double w[N], v[N];
	cout << "输入物品种数:" << endl;
	cin >> n;

	cout << "输入限制重量:" << endl;
	cin >> limitW;
	for (k = 0; k < n; ++k) {
		cout << "依次输入第" << k + 1 << "个物品的重量和价值: " << endl;
		cin >> w[k] >> v[k];
		a[k].weight = w[k];
		a[k].value = v[k];
	}

	maxv = 0.0;
	for (k = 0; k < n; ++k)cop[k] = 0;
	BackTrack(0, 0.0, 0.0);
	cout << "所选物品为：" << endl;
	for (k = 0; k < n; ++k)
		if (option[k])
			cout << k + 1 << "\t";
	cout << endl << "总价值为:" << maxv << endl;
}

动态规划

#include<iostream>
using namespace std;
int weight[100], goods_value[100];//weight表示背包可容纳重量，godds_value表示物品价值
int Value[100][100];//价值数组
int ZeroOneBag(int goods_value[], int weight[], int Value[][100], int m, int n) {
	for (int i = 1; i <= m; i++) {
		for (int j = 1; j <= n; j++) {
			if (j < weight[i]) {
				Value[i][j] = Value[i - 1][j];
			} else if (Value[i - 1][j] >= (Value[i - 1][j - weight[i]] + goods_value[i])) {
				Value[i][j] = Value[i - 1][j];
			} else {
				Value[i][j] = Value[i - 1][j - weight[i]] + goods_value[i];
			}
		}
	}
	return Value[m][n];
}
int main() {

	int i = 0, m, n, value = 0;
	cout << "请输入物品个数：" << endl;
	cin >> m;
	cout << "请输入背包容量：\n";
	cin >> n;
	for (i = 1; i <= m; i++) {
		cout << "请输入物品" << i << endl;
		cout << "重量: " << "价值: \n";
		cin >> weight[i] >> goods_value[i];

	}

	value = ZeroOneBag(goods_value, weight, Value, m, n);
	cout << "背包能容纳的最大价值为：" << value << endl;
	cout << endl;
}

贪心

#include <iostream>
#include<algorithm>
using namespace std;
struct goodinfo {
	float p; //物品效益
	float w; //物品重量
	float X; //物品该放的数量
	int flag; //物品编号
	bool operator<(goodinfo b) {
		return p > b.p;
	}//按物品效益，重量比值做升序排列
};//物品信息结构体
void bag(goodinfo goods[], float M, int n) {
	float cu;
	int i, j;
	for (i = 1; i <= n; i++)
		goods[i].X = 0;
	cu = M; //背包剩余容量
	for (i = 1; i < n; i++) {
		if (goods[i].w > cu) //当该物品重量大与剩余容量跳出
			continue;
		goods[i].X = 1;
		cu = cu - goods[i].w; //确定背包新的剩余容量
	}
//按物品编号做降序排列
	for (j = 2; j <= n; j++) {
		goods[0] = goods[j];
		i = j - 1;
		while (goods[0].flag < goods[i].flag) {
			goods[i + 1] = goods[i];
			i--;
		}
		goods[i + 1] = goods[0];
	}
	cout << "最优解为:" << endl;
	double ans = 0;
	for (i = 1; i <= n; i++) {
		cout << "第" << i << "件物品要放:";
		cout << goods[i].X << endl;
		ans += goods[i].p * goods[i].w * goods[i].X;
	}
	cout << "总价值为：" << ans << endl;

}
int main(void) {
	cout << "|--------运用贪心法解背包问题---------|" << endl;
	cout << "|-------------------------------------|" << endl;
	int i, n;
	float M;
	cout << "请输入物品的总数量：";
	cin >> n;
	cout << "请输入背包的最大容量：";
	cin >> M;
	goodinfo goods[n + 1];

	cout << endl;
	for (i = 1; i <= n; i++) {
		goods[i].flag = i;
		cout << "请输入第" << i << "件物品的重量:";
		cin >> goods[i].w;
		cout << "请输入第" << i << "件物品的效益:";
		cin >> goods[i].p;
		goods[i].p = goods[i].p / goods[i].w; //得出物品的效益，重量比
		cout << endl;
	}

	sort(goods + 1, goods + 1 + n);
	bag(goods, M, n);

	return 0;
}

模拟退火

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
void knapsackSa(int w[], int c[], int n, int M) { //n件物品，其重量和价值分别为w[i]和c[i]，寻找将其装入容量为M的背包中物品的最大价值
	int i, j, df, dm;
	int *x = new int[n]; //定义解空间
	for (i = 0; i < n; i++) { //初始化解为0
		x[i] = 0;
	}
	int f = 0, m = 0;
	for (i = 0; i < n; i++) {
		f = f + c[i] * x[i]; //初始化总价值
		m = m + w[i] * x[i];	 //初始化总重量
	}
	float t0 = 500; //控制参数t的初值
	float t = t0;
	float a = 0.95f; //衰减函数的系数
	float e = 0.00001f;
	int L = 100 * n; // Mapkob 链长
	while (t > e) { //停止准则
		srand((unsigned)time(NULL));//初始化随机函数种子，srand((unsigned)time(NULL));是拿系统时间作为种子，由于时间是变化的，种子变化，可以产生不相同的随机数。
		for (int k = 0; k < L; k++) {
			i = rand() % n; //随机选取第i件物品
			if (x[i] == 0) {	//若i不在背包中
				if (m + w[i] <= M) {	//且加入总重量后不超过容量M,则直接放入背包中
					x[i] = 1;
					f = f + c[i];
					m = m + w[i];
				} else {
					j = rand() % n;	//随机取出物品j
					while (x[j] == 0) {
						j = rand() % n;   //直到x[j]为1
					}
					df = c[i] - c[j];
					dm = w[i] - w[j];
					if (m + dm <= M)	//加入总重量后不超过容量M
						if (df > 0 || (exp(df / t) > (double)(rand() / (double)RAND_MAX))) { //价值差大于0或以exp(df/T)的接受概率接受新解
							x[i] = 1;
							x[j] = 0;
							f = f + df;
							m = m + dm;
						}
				}
			} else {
				j = rand() % n;
				while (x[j] == 1) {
					j = rand() % n;
				}
				df = c[j] - c[i];
				dm = w[j] - w[i];
				if (m + dm <= M)
					if (df > 0 || (exp(df / t) > (double)(rand() / (double)RAND_MAX))) { //价值差大于0或以exp(df/T)的接受概率接受新解
						x[i] = 0;
						x[j] = 1;
						f = f + df;
						m = m + dm;
					}
			}
		}
		t = t * a; //衰减函数
	}
	cout << "该0/1背包问题的最优解为： ";
	for (i = 0; i <= n - 1; i++) cout << x[i] << " ";
	cout << endl << "最大总价值为：" << f << endl;
}
int main() {
	int n, M;
	//n件物品，其重量和价值分别为w[i]和c[i]，寻找将其装入容量为M的背包中物品的最大价值
	cout << "请输入物品件数n:" << endl;
	cin >> n;
	cout << "请输入背包容量M:" << endl;
	cin >> M;
	int *w = new int[n];
	cout << "请依次输入物品重量和价值:" << endl;
	int *c = new int[n];
	for (int i = 0; i < n; i++) {
		cin >> w[i] >> c[i];
	}

	knapsackSa(w, c, n, M);
	return 0;
}

测试用例文章来源地址https://www.toymoban.com/news/detail-784892.html