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229 lines (185 loc) · 6.68 KB
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import sys
sys.setrecursionlimit(10000)
def unit_propagation(clauses, assignment):
queue = []
for c in clauses:
if len(c) == 1:
queue.append(c[0])
while queue:
lit = queue.pop()
var = abs(lit)
val = lit > 0
if var in assignment:
if assignment[var] != val:
return None
continue
assignment[var] = val
new_clauses = []
for c in clauses:
if lit in c:
continue
if -lit in c:
new_c = [x for x in c if x != -lit]
if len(new_c) == 0:
return None
if len(new_c) == 1:
queue.append(new_c[0])
new_clauses.append(new_c)
else:
new_clauses.append(c)
clauses = new_clauses
return clauses
def choose_variable(clauses, assignment):
for c in clauses:
for lit in c:
var = abs(lit)
if var not in assignment:
return var
return None
def dpll(clauses, assignment):
clauses = unit_propagation(clauses, assignment)
if clauses is None:
return None
if not clauses:
return assignment
var = choose_variable(clauses, assignment)
if var is None:
return assignment
for val in [True, False]:
new_assignment = assignment.copy()
new_assignment[var] = val
lit = var if val else -var
new_clauses = []
for c in clauses:
if lit in c:
continue
if -lit in c:
new_c = [x for x in c if x != -lit]
if len(new_c) == 0:
break
new_clauses.append(new_c)
else:
new_clauses.append(c)
else:
result = dpll(new_clauses, new_assignment)
if result is not None:
return result
return None
def main():
input = sys.stdin.readline
n, m = map(int, input().split())
clauses = [list(map(int, input().split())) for _ in range(m)]
result = dpll(clauses, {})
if result is None:
print("UNSAT")
else:
print("SAT")
print(" ".join('1' if result.get(i, False) else '0' for i in range(1, n+1)))
if __name__ == "__main__":
main()
import sys
# Повышаем лимит рекурсии для глубокого дерева DPLL
sys.setrecursionlimit(2000)
class SAT_Solver:
def __init__(self, num_vars, clauses):
self.num_vars = num_vars
self.clauses = clauses
self.assignment = [0] * (num_vars + 1) # 0: undef, 1: True, -1: False
def unit_propagate(self):
"""Реализация продвижения единичных литералов из Лекции 9."""
changed = True
while changed:
changed = False
for clause in self.clauses:
unassigned = []
satisfied = False
for lit in clause:
var = abs(lit)
val = 1 if lit > 0 else -1
if self.assignment[var] == val:
satisfied = True
break
if self.assignment[var] == 0:
unassigned.append(lit)
if satisfied:
continue
if not unassigned:
return False # Конфликт
if len(unassigned) == 1:
lit = unassigned[0]
self.assignment[abs(lit)] = 1 if lit > 0 else -1
changed = True
return True
def solve(self):
"""Классический алгоритм DPLL с возвратами."""
if not self.unit_propagate():
return False
# Выбираем следующую неназначенную переменную
var = 0
for i in range(1, self.num_vars + 1):
if self.assignment[i] == 0:
var = i
break
x
if var == 0:
return True # Решение найдено
state = list(self.assignment)
# Пробуем True
self.assignment[var] = 1
if self.solve():
return True
# Откат и проба False
self.assignment = state
self.assignment[var] = -1
if self.solve():
return True
return False
# Чтение входных данных
line1 = sys.stdin.readline().split()
n, l_bag, w_bag = map(int, line1)
lengths = list(map(int, sys.stdin.readline().split()))
# 1. Генерация возможных позиций (i, row, col)
placements = []
box_vars = [[] for _ in range(n)]
for i in range(n):
box_len = lengths[i]
for r in range(1, w_bag + 1):
for c in range(1, l_bag - box_len + 2):
var_idx = len(placements) + 1
placements.append({'id': i, 'r': r, 'c': c, 'len': box_len})
box_vars[i].append(var_idx)
clauses = []
# 2. Формирование ограничений КНФ
# Каждая пачка должна быть хотя бы в одном месте
for i in range(n):
if not box_vars[i]:
print("UNSAT")
clauses.append(box_vars[i])
# Пачка не может быть в двух местах одновременно
for i in range(n):
vars_i = box_vars[i]
for idx_a in range(len(vars_i)):
for idx_b in range(idx_a + 1, len(vars_i)):
clauses.append([-vars_i[idx_a], -vars_i[idx_b]])
# Проверка перекрытий пачек в багажнике
for a in range(len(placements)):
for b in range(a + 1, len(placements)):
p1, p2 = placements[a], placements[b]
if p1['id'] == p2['id']: continue
if p1['r'] == p2['r']:
# Проверка пересечения отрезков в одном ряду
if max(p1['c'], p2['c']) < min(p1['c'] + p1['len'], p2['c'] + p2['len']):
clauses.append([-(a + 1), -(b + 1)])
# 3. Запуск решателя
solver = SAT_Solver(len(placements), clauses)
if solver.solve():
print("SAT")
results = [None] * n
for i, val in enumerate(solver.assignment[1:]):
if val == 1:
p = placements[i]
results[p['id']] = (p['r'], p['c'])
for res in results:
print(f"{res[0]} {res[1]}")
else:
print("UNSAT")