笔试-士兵过河

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 c1
    else:
        return c2

i = 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 + 1
            j = j + 1
        else:
            # 下一次循环的索引
            j = j + 1
            continue
    else:
        # 下一次循环的索引
        i = i + 1
        continue
    
    if 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