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应用

一个军队有N名士兵需要过河，敌军在T时长后追上。现有一条小船，最多乘坐2名士兵。
当一名士兵划船过河时，用时为a[i]，0<=i<N；
当两名士兵共同划船过河时，用时为max(a[i], a[j])，两士兵中单独划船较长的；
当两名士兵乘船、一名士兵划船过河时，用时为10*a[i]，a[i]为单独划船的士兵；
士兵游泳过河直接标记为死亡。
#请设计一种方案，保证存活的士兵尽可能多、过河用时尽可能短。

实现

N = int(input("士兵数："))
T = int(input("敌军到达时长："))
strings = input("每个士兵过河时长：").split()
a = [int(i) for i in strings]# 回
back_solution = []
for i in range(0, len(a)):g = [i, a[i]]back_solution.append(g)
# 去1
to_solution_1 = []
for i in range(0, len(a)-1):for j in range(i+1, len(a)):g = [i, j, max(a[i], a[j])]to_solution_1.append(g)
# 去2
to_solution_2 = []    
for i in range(0, len(a)):for j in range(0, len(a)):if i != j:# i为划船士兵g = [i, j, 10*a[i]]to_solution_2.append(g)# print(f"回：{back_solution}")
# print(f"去一：{to_solution_1}")
# print(f"去二：{to_solution_2}")# 为保证有效运输，采取“去2人回1人”，不能是“去1人回1人”、“去2人回2人”、“去2人回0人”。去+回=周期
cycles1 = []
for i in range(0, len(to_solution_1)):for j in range(0, len(back_solution)):driver0 = to_solution_1[i][0]driver1 = to_solution_1[i][1]back_member = back_solution[j][0]if back_member == driver0 or back_member == driver1:time = to_solution_1[i][-1] + back_solution[j][-1]g = [driver0, driver1, back_member, time]cycles1.append(g)cycles2 = []
for i in range(0, len(to_solution_2)):for j in range(0, len(back_solution)):driver0 = to_solution_2[i][0]driver1 = to_solution_2[i][1]back_member = back_solution[j][0]if back_member == driver0 or back_member == driver1:time = to_solution_2[i][-1] + back_solution[j][-1]g = [driver0, driver1, back_member, time]cycles2.append(g)# 从小到大排序
for i in range(0, len(cycles1)):for j in range(0, len(cycles1)):if cycles1[i][-1] < cycles1[j][-1]:cycles1[j], cycles1[i] = cycles1[i], cycles1[j]for i in range(0, len(cycles2)):for j in range(0, len(cycles2)):if cycles2[i][-1] < cycles2[j][-1]:cycles2[j], cycles2[i] = cycles2[i], cycles2[j]# print(f"组合一：{cycles1}")
# print(f"组合二：{cycles2}")def compare(c1, c2):if c1[-1] <= c2[-1]:return c1else:return c2i = 0
j = 0
safe = []
time = []
while T > 0:# 当前组合的乘船士兵，不包含已到达对岸的士兵if (cycles1[i][0] not in safe) and (cycles1[i][1] not in safe):if (cycles2[j][0] not in safe) and (cycles2[j][1] not in safe):# 最短时间方案c = compare(cycles1[i], cycles2[j])# 下一次循环的索引i = i + 1j = j + 1else:# 下一次循环的索引j = j + 1continueelse:# 下一次循环的索引i = i + 1continueif c[0] == c[2]:safe.append(c[1])else:safe.append(c[0])time.append(c[-1])T = T - c[-1]# print(f"临界值：{i-1}或{j-1}")
c = compare(cycles1[i-1], cycles2[j-1])
bak_index = c[2]
safe.append(bak_index)sum = 0
for i in time:sum = sum + i
t = sum - a[bak_index]
print(len(safe), t)

士兵数：5
敌军到达时长：43
每个士兵过河时长：12 13 15 20 50
3 40