利用贪心算法解决背包问题

#include <stdio.h>
#include<stdlib.h>
#define MAXSIZE 100 //假设物体总数
#define M 20 //背包的载荷能力

//算法核心，
void GREEDY(float w[], float x[], int sortResult[], int n)
{
float cu = M;
int i = 0;
int temp = 0;

for (i = 0; i < n; i++)//准备输出结果
{
x[i] = 0;
}

for (i = 0; i < n; i++)
{
temp = sortResult[i];//得到取物体的顺序
if (w[temp] > cu)
{
break;
}

x[temp] = 1;//若合适则取出
cu -= w[temp];//将容量相应的改变
}

if (i <= n)//使背包充满
{
x[temp] = cu / w[temp];
}

return;
}

void sort(float tempArray[], int sortResult[], int n)
{
int i = 0, j = 0;
int index = 0, k = 0;

for (i = 0; i < n; i++)//对映射数组赋初值0
{
sortResult[i] = 0;
}

for (i = 0; i < n; i++)
{
float temp = tempArray[i];

index = i;

//找到最大的效益并保存此时的下标
for (j = 0; j < n; j++)
{
if ((temp < tempArray[j]) && (sortResult[j] == 0))
{
temp = tempArray[j];
index = j;
}
}

//对w[i]作标记排序
if (sortResult[index] == 0)
{
sortResult[index] = ++k;
}
}

//修改效益最低的sortResult[i]标记
for (i = 0; i < n; i++)
{
if (sortResult[i] == 0)
{
sortResult[i] = ++k;
}
}

return;
}

//得到本算法的所有输入信息
void getData(float p[], float w[], int *n)
{
int i = 0;

printf("please input the total count of object: ");
scanf("%d", n);

printf("Please input array of p :\n");
for (i = 0; i < (*n); i++)
{
scanf("%f", &p[i]);
}

printf("Now please input array of w :\n");
for (i = 0; i < (*n); i++)
{
scanf("%f", &w[i]);
}

return;
}

void output(float x[], int n)
{
int i;

printf("\n\nafter arithmetic data: advise method\n");
for (i = 0; i < n; i++)
{
printf("x[%d]\t", i);
}

printf("\n");
for (i = 0; i < n; i++)
{
printf("%2.3f\t", x[i]);
}

return;
}

int main()
{
float p[MAXSIZE], w[MAXSIZE], x[MAXSIZE];
int i = 0, n = 0;
int sortResult[MAXSIZE];

getData(p, w, &n);

for (i = 0; i < n; i++)
{
x[i] = p[i] / w[i];
}

sort(x, sortResult, n);

GREEDY(w, x, sortResult, n);

output(x, n);
system("pause");

}