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satisfiability.cc
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112 lines (86 loc) · 2.34 KB
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// William Sjöblom
// Nasty brute force approach but does the job.
// Very much possible to eliminate recursion branches.
#include <cstdio>
#include <cmath>
#include <vector>
struct Clause {
std::vector<int> literals;
void dump() {
if (literals[0] < 0) printf("~");
printf("X%d", std::abs(literals[0]));
for (int i = 1; i < literals.size(); i++) {
printf(" v ");
if (literals[i] < 0) printf("~");
printf("X%d", std::abs(literals[i]));
}
printf("\n");
}
bool satisfied(std::vector<bool>& values) {
bool satisfied = false;
for (int literal : literals) {
satisfied |= (literal < 0) ^ values[std::abs(literal) - 1];
}
return satisfied;
}
};
/**
* Scan variable.
*/
int scan_var() {
int var;
if(scanf("~X%d", &var)) {
var = -var;
} else {
scanf("X%d", &var);
}
return var;
}
/**
* Scan clause.
*/
Clause scan_clause() {
Clause c;
c.literals.push_back(scan_var());
while (getchar() == ' ') { // Will eventually consume a newline and break loop.
scanf("v ");
c.literals.push_back(scan_var());
}
return c;
}
bool solve0(std::vector<Clause>& clauses, std::vector<bool> values, int i=0) {
if (i >= values.size()) {
for (Clause& clause : clauses) {
if (!clause.satisfied(values)) return false;
}
return true;
}
bool first = solve0(clauses, values, i + 1);
if (first) return true;
std::vector<bool> other = values;
other[i] = !other[i];
return solve0(clauses, other, i + 1);
}
bool solve(int var_count, std::vector<Clause>& clauses) {
std::vector<bool> values(var_count);
return solve0(clauses, values, 0);
}
/**
* Entry point.
*/
int main() {
int test_count; scanf("%d", &test_count);
while (test_count--) {
int var_count, clause_count; scanf("%d %d\n", &var_count, &clause_count);
std::vector<Clause> clauses;
clauses.reserve(clause_count);
while (clause_count--) {
Clause c = scan_clause();
clauses.push_back(c);
}
bool satisfiable = solve(var_count, clauses);
if (satisfiable) printf("satisfiable\n");
else printf("unsatisfiable\n");
}
return 0;
}