今天做作业，要实现整数线性规划的分枝定界法算法。找了一些网上的博客，发现都很屎，感觉自己写的这个比较清楚、规范，所以在此记录。如有错误，请指正。

from scipy.optimize import linprog
import numpy as np
import math
import sys
from queue import Queueclass ILP():def __init__(self, c, A_ub, b_ub, A_eq, b_eq, bounds):# 全局参数self.LOWER_BOUND = -sys.maxsizeself.UPPER_BOUND = sys.maxsizeself.opt_val = Noneself.opt_x = Noneself.Q = Queue()# 这些参数在每轮计算中都不会改变self.c = -cself.A_eq = A_eqself.b_eq = b_eqself.bounds = bounds# 首先计算一下初始问题r = linprog(-c, A_ub, b_ub, A_eq, b_eq, bounds)# 若最初问题线性不可解if not r.success:raise ValueError('Not a feasible problem!')# 将解和约束参数放入队列self.Q.put((r, A_ub, b_ub))def solve(self):while not self.Q.empty():# 取出当前问题res, A_ub, b_ub = self.Q.get(block=False)# 当前最优值小于总下界，则排除此区域if -res.fun < self.LOWER_BOUND:continue# 若结果 x 中全为整数，则尝试更新全局下界、全局最优值和最优解if all(list(map(lambda f: f.is_integer(), res.x))):if self.LOWER_BOUND < -res.fun:self.LOWER_BOUND = -res.funif self.opt_val is None or self.opt_val < -res.fun:self.opt_val = -res.funself.opt_x = res.xcontinue# 进行分枝else:# 寻找 x 中第一个不是整数的，取其下标 idxidx = 0for i, x in enumerate(res.x):if not x.is_integer():breakidx += 1# 构建新的约束条件（分割new_con1 = np.zeros(A_ub.shape[1])new_con1[idx] = -1new_con2 = np.zeros(A_ub.shape[1])new_con2[idx] = 1new_A_ub1 = np.insert(A_ub, A_ub.shape[0], new_con1, axis=0)new_A_ub2 = np.insert(A_ub, A_ub.shape[0], new_con2, axis=0)new_b_ub1 = np.insert(b_ub, b_ub.shape[0], -math.ceil(res.x[idx]), axis=0)new_b_ub2 = np.insert(b_ub, b_ub.shape[0], math.floor(res.x[idx]), axis=0)# 将新约束条件加入队列，先加最优值大的那一支r1 = linprog(self.c, new_A_ub1, new_b_ub1, self.A_eq,self.b_eq, self.bounds)r2 = linprog(self.c, new_A_ub2, new_b_ub2, self.A_eq,self.b_eq, self.bounds)if not r1.success and r2.success:self.Q.put((r2, new_A_ub2, new_b_ub2))elif not r2.success and r1.success:self.Q.put((r1, new_A_ub1, new_b_ub1))elif r1.success and r2.success:if -r1.fun > -r2.fun:self.Q.put((r1, new_A_ub1, new_b_ub1))self.Q.put((r2, new_A_ub2, new_b_ub2))else:self.Q.put((r2, new_A_ub2, new_b_ub2))self.Q.put((r1, new_A_ub1, new_b_ub1))def test1():""" 此测试的真实最优解为 [4, 2] """c = np.array([40, 90])A = np.array([[9, 7], [7, 20]])b = np.array([56, 70])Aeq = Nonebeq = Nonebounds = [(0, None), (0, None)]solver = ILP(c, A, b, Aeq, beq, bounds)solver.solve()print("Test 1's result:", solver.opt_val, solver.opt_x)print("Test 1's true optimal x: [4, 2]\n")def test2():""" 此测试的真实最优解为 [2, 4] """c = np.array([3, 13])A = np.array([[2, 9], [11, -8]])b = np.array([40, 82])Aeq = Nonebeq = Nonebounds = [(0, None), (0, None)]solver = ILP(c, A, b, Aeq, beq, bounds)solver.solve()print("Test 2's result:", solver.opt_val, solver.opt_x)print("Test 2's true optimal x: [2, 4]\n")if __name__ == '__main__':test1()test2()

运行结果截图：