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#include <regex.h>
#include <stdio.h>
#include <stdlib.h>
#define arrlen(x) (sizeof(x)/sizeof((x)[0]))
enum MATERIAL {
ORE,
CLAY,
BSDN,
GEOD,
};
struct blueprint {
int id;
int ore_ore;
int clay_ore;
int bsdn_ore;
int bsdn_clay;
int geod_ore;
int geod_bsdn;
};
void
blueprint_ctor(struct blueprint * b, int id, int ore_ore, int clay_ore,
int bsdn_ore, int bsdn_clay, int geod_ore, int geod_bsdn)
{
b->id = id;
b->ore_ore = ore_ore;
b->clay_ore = clay_ore;
b->bsdn_ore = bsdn_ore;
b->bsdn_clay = bsdn_clay;
b->geod_ore = geod_ore;
b->geod_bsdn = geod_bsdn;
}
struct material {
long amount;
long robots;
long pending;
};
void
material_ctor(struct material * m, long amount, long robots)
{
m->amount = amount;
m->robots = robots;
m->pending = 0;
}
// Tick a material.
//
// Amount increases by the number of robots
// Number of robots increases by number of pending robots
void
material_tick(struct material * m)
{
m->amount += m->robots;
m->robots += m->pending;
m->pending = 0;
}
struct inventory {
struct material ore;
struct material clay;
struct material bsdn;
struct material geod;
};
void
inventory_init(struct inventory * i)
{
material_ctor(&i->ore, 0, 1);
material_ctor(&i->clay, 0, 0);
material_ctor(&i->bsdn, 0, 0);
material_ctor(&i->geod, 0, 0);
}
void
inventory_tick(struct inventory * i)
{
material_tick(&i->ore);
material_tick(&i->clay);
material_tick(&i->bsdn);
material_tick(&i->geod);
}
// Try to build a robot out of material #m
int
inventory_build(struct inventory * i, struct blueprint const * b, enum MATERIAL m)
{
switch (m) {
case ORE:
if (i->ore.amount < b->ore_ore)
return -1;
i->ore.amount -= b->ore_ore;
i->ore.pending += 1;
break;
case CLAY:
if (i->ore.amount < b->clay_ore)
return -1;
i->ore.amount -= b->clay_ore;
i->clay.pending += 1;
break;
case BSDN:
if (i->ore.amount < b->bsdn_ore
|| i->clay.amount < b->bsdn_clay)
return -1;
i->ore.amount -= b->bsdn_ore;
i->clay.amount -= b->bsdn_clay;
i->bsdn.pending += 1;
break;
case GEOD:
if (i->ore.amount < b->geod_ore
|| i->bsdn.amount < b->geod_bsdn)
return -1;
i->ore.amount -= b->geod_ore;
i->bsdn.amount -= b->geod_bsdn;
i->geod.pending += 1;
break;
default:
return -1;
}
return 0;
}
// Most basic build algorithm. Build as many of the most important things that you
// can, and then as many of the next most important, and so on.
//
// Not optimal. But it helps get the structure of the program working.
void
inventory_build_basic(struct inventory * i, struct blueprint const * b, int remaining)
{
int t;
for (t = 0; t < remaining; ++t) {
while (inventory_build(i, b, GEOD) == 0)
;
while (inventory_build(i, b, BSDN) == 0)
;
while (inventory_build(i, b, CLAY) == 0)
;
while (inventory_build(i, b, ORE) == 0)
;
inventory_tick(i);
fprintf(stderr, "Blueprint %d tick %d, ore: %ld/%ld, clay %ld/%ld, bsdn %ld/%ld, geod %ld/%ld\n",
b->id, t + 1,
i->ore.amount, i->ore.robots,
i->clay.amount, i->clay.robots,
i->bsdn.amount, i->bsdn.robots,
i->bsdn.amount, i->bsdn.robots);
}
}
int
main()
{
regex_t reblueprint;
char buf[BUFSIZ];
if (regcomp(&reblueprint, "Blueprint ([[:digit:]]+):"
" Each ore robot costs ([[:digit:]]+) ore."
" Each clay robot costs ([[:digit:]]+) ore."
" Each obsidian robot costs ([[:digit:]]+) ore and ([[:digit:]]+) clay."
" Each geode robot costs ([[:digit:]]+) ore and ([[:digit:]]+) obsidian.",
REG_EXTENDED) != 0)
{
fprintf(stderr, "Bad regex\n");
return -1;
}
while (fgets(buf, sizeof(buf), stdin)) {
regmatch_t rematch[8];
struct blueprint b;
struct inventory i;
if (regexec(&reblueprint, buf, arrlen(rematch), rematch, 0) != 0) {
fprintf(stderr, "Unexpected blueprint in %s\n", buf);
continue;
}
blueprint_ctor(&b,
atoi(buf + rematch[1].rm_so),
atoi(buf + rematch[2].rm_so),
atoi(buf + rematch[3].rm_so),
atoi(buf + rematch[4].rm_so),
atoi(buf + rematch[5].rm_so),
atoi(buf + rematch[6].rm_so),
atoi(buf + rematch[7].rm_so));
inventory_init(&i);
inventory_build_basic(&i, &b, 24);
printf("Blueprint %d geodes = %ld\n", b.id, i.geod.amount);
}
regfree(&reblueprint);
return 0;
}
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