算法之美：0-1背包问题（动态规划法，回溯法，贪心法）

#include <stdio.h>

#define CAPACITY  10                        //背包的容量
#define N         5                         //n为物品的个数

int max(int a, int b)
{
   return a > b ? a : b;
}

void print_array(int *v, int n)
{
    int i;

    for (i = 0; i < n; ++i)
    {
        printf("%3d ", v[i]);
    }
    printf("\n");
}

int package0_1(int m[][CAPACITY + 1], int *wt, int *value, int n)//n代表物品的个数
{
    int i = 0;					//个数
    int w;						//重量

    /*********************************放置前0个物品*********************************/
    for (w = 0; w <= CAPACITY; w++)
    {
        m[i][w] = 0;
    }

    /*********************************放置1 ~ n个物品**********************************/
    for(i = 1; i <= n; i++)
    {
        for(w = 0; w <= CAPACITY; w++)
        {
            if (wt[i - 1] <= w)
            {
                    m[i][w] = max(value[i-1] + m[i-1][w-wt[i-1]], m[i-1][w]);
            }
            else
            {
                    m[i][w] = m[i-1][w];
            }
        }
    }

	return m[n][CAPACITY];
}

void answer( int *x, int m[][CAPACITY + 1], int *wt, int n)
{
    int w = CAPACITY;                       /*i = n, j= CAPACITY坐标上存放着背包容量为c时的最大价值*/
    int i;

    for(i = n - 1; i >= 0; i--)
    {
        if(m[i + 1][w] == m[i][w])
        {
            x[i] = 0;
        }
        else
        {
            x[i] = 1;                 /*如果当前物品放入了背包*/
            w = w - wt[i];            /*重新计算背包剩余容量，以计算在取最优时其他物品的取舍情况*/
        }
    }
}

int main()
{
    int wt[N]    = {2, 2, 6, 5, 4};             //物品的重量
    int value[N] = {2, 3, 5, 4, 6};             //物品对应的价值
    int m[N + 1][CAPACITY + 1] = {0}; 			//动态规划表， 行号代表选择几件放入保证，列号表示装入物品的总重量
    int x[N] = {0};								//答案表，每一个物品是否放入包中
    int i;

	printf("The best value is: %d\n", package0_1(m, wt, value, N));

    for(i = 0; i <= CAPACITY; i++)
    {
        printf("%3d ", i);
    }
    printf("\n");

    for(i = 0; i <= N; i++)
    {
        print_array(m[i], CAPACITY + 1);
    }

    answer(x, m, wt, N);

    printf("The best answer is:\n");
    print_array(x, N);

    return 0;
}

#include <stdio.h>
 
#define N 5         //物品的数量
#define C  10       //背包的容量
 
int w[N] = {2, 2, 6, 5, 4};  //每个物品的重量
int v[N] = {6, 3, 5, 4, 6};   //每个物品的价值
int x[N] = {0, 0, 0, 0, 0};   //x[i]=1代表物品i放入背包，0代表不放入
 
int cur_weight = 0;  //当前放入背包的物品总重量
int cur_value = 0;   //当前放入背包的物品总价值
 
int best_value = 0;  //最优值；当前的最大价值，初始化为0
int best_x[N];       //最优解；best_x[i]=1代表物品i放入背包，0代表不放入
 
//t = 0 to N-1
void backtrack(int t)
{
	int i;

	//叶子节点，输出结果
	if(t > N - 1) 
	{
		//如果找到了一个更优的解
		if(cur_value > best_value)
		{
			//保存更优的值和解
			best_value = cur_value;
			for(i = 0; i < N; ++i) 
			{
				best_x[i] = x[i];
			}
		}
	}
	else
	{
		//遍历当前节点的子节点：0 不放入背包，1放入背包
		for(i = 0; i <= 1; ++i)
		{
			x[t] = i;
 
			if(i == 0)         //不放入背包
			{
				backtrack(t + 1);
			}
			else               //放入背包
			{
 				//约束条件：放的下
				if((cur_weight + w[t]) <= C)
				{
					cur_weight += w[t];
					cur_value += v[t];
					backtrack(t + 1);
					cur_weight -= w[t];
					cur_value -= v[t];
				}
			}
		}
	}
}
 
 //回溯法解决0-1背包问题
int main(int argc, char* argv[])
{
	int i; 
	
	backtrack(0);
 
	printf("最优值：%d\n",best_value);
 
	for(i = 0; i < N; i++)
	{
	   printf("%-3d", best_x[i]);
	}

	return 0;
}

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

typedef struct good_info
{
    double price;      //物品效益
    double sum_price;  //物品总价值
    double get_rate;   //获得此物品占总数的比例
    int get_weight;    //获得此物品的物品数量
    int sum_weight;    //物品总重量
}good_info;

 void print_array(good_info *goods, int n);

/*********按物品效益，重量比值做序排列***************/
void insert_sort_by_price(good_info *goods, int n)
{
    int i, j;
    good_info tmp;

    for(i = 1; i < n; i++)
    {

        tmp = goods[i];
        j = i - 1;
        while ((j >= 0) && (tmp.price > goods[j].price))
        {
            goods[j + 1] = goods[j];
            j--;
        }

        goods[j + 1] = tmp;
    }
}

 void bag(good_info *goods, int capacity, int n)
 {
 	int left;
    int i;

    for(i = 0; i < n; i++)
    {
    	goods[i].get_weight = 0;
    }

    left = capacity;                     //背包剩余容量

    for(i = 0; i < n; i++)
    {
        if(left <= goods[i].sum_weight )   //当该物品重量大与剩余容量跳出
        {
            break;
       	}

        goods[i].get_weight = goods[i].sum_weight;
        goods[i].get_rate  = 1.0;

        left -= goods[i].sum_weight;                        //确定背包新的剩余容量
    }

    if(i < n)
    {
    	goods[i].get_weight = left;                          //该物品所要放的量
    	goods[i].get_rate = left * 1.0 / goods[i].sum_weight;//该物品所要放的量
    }

 }

 void print_array(good_info *goods, int n)
 {
 	int i;

 	for(i = 0; i < n; i++)
	{
		printf("%d\t%lf\t%lf\t%d\t%lf\n",
				goods[i].sum_weight,
				goods[i].sum_price,  goods[i].price,
				goods[i].get_weight, goods[i].get_rate);
	}

}

//贪心法解决背包问题
int main(int argc, char const *argv[])
{
	int n = 10;
    int i;
    int  capacity;

    good_info *goods;//定义一个指针
    goods = (good_info *)malloc(sizeof(good_info) * n);
	if (goods == NULL)
	{
		printf("malloc failed\n");
		exit(1);
	}

    srand(time(NULL));
   	for(i = 0; i < n; i++)
	{
		goods[i].sum_weight = rand() % 50 +1;
		goods[i].sum_price = rand() % 100 + 1;
		goods[i].price = goods[i].sum_price/goods[i].sum_weight;//得出物品的重量比
	}

	printf("背包的最大容量：");
    scanf("%d", &capacity);

    insert_sort_by_price(goods, n);
	bag(goods, capacity, n);
   	print_array(goods, n);

	free(goods);

	return 0;
}