programming-examples/c/Numerical/C Program to Solve the 0-1 Knapsack Problem.c
2019-11-15 12:59:38 +01:00

95 lines
2.4 KiB
C

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
typedef struct
{
const char * name;
int weight, value;
} item_t;
item_t item[] =
{
{"map", 9, 150},
{"compass", 13, 35},
{"water", 153, 200},
{"sandwich", 50, 160},
{"glucose", 15, 60},
{"tin", 68, 45},
{"banana", 27, 60},
{"apple", 39, 40},
{"cheese", 23, 30},
{"beer", 52, 10},
{"suntancream", 11, 70},
{"camera", 32, 30},
{"T-shirt", 24, 15},
{"trousers", 48, 10},
{"umbrella", 73, 40},
{"waterproof trousers", 42, 70},
{"waterproof overclothes", 43, 75},
{"note-case", 22, 80},
{"sunglasses", 7, 20},
{"towel", 18, 12},
{"socks", 4, 50},
{"book", 30, 10}
};
#define n_items (sizeof(item)/sizeof(item_t))
typedef struct
{
uint32_t bits; /* 32 bits, can solve up to 32 items */
int value;
} solution;
void optimal(int weight, int idx, solution *s)
{
solution v1, v2;
if (idx < 0)
{
s->bits = s->value = 0;
return;
}
if (weight < item[idx].weight)
{
optimal(weight, idx - 1, s);
return;
}
optimal(weight, idx - 1, &v1);
optimal(weight - item[idx].weight, idx - 1, &v2);
v2.value += item[idx].value;
v2.bits |= (1 << idx);
*s = (v1.value >= v2.value) ? v1 : v2;
}
int main(void)
{
int i = 0, w = 0;
solution s = {0, 0};
optimal(400, n_items - 1, &s);
for (i = 0; i < n_items; i++)
{
if (s.bits & (1 << i))
{
printf("%s\n", item[i].name);
w += item[i].weight;
}
}
printf("Total value: %d; weight: %d\n", s.value, w);
return 0;
}
/*
map
compass
water
sandwich
glucose
banana
suntancream
waterproof trousers
waterproof overclothes
note-case
sunglasses
socks
Total value: 1030; weight: 396