洛谷\P1049 [NOIP2001 普及组] 装箱问题\动态规划\Python

链接：P1049 [NOIP2001 普及组] 装箱问题 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#题解1
#dp[i][j]代表前i个物品，容量为j的情况下的最大容积
#dp[i][j] = max(dp[i-1][j],dp[i-1][j-lst[i]] + lst[i])
v = int(input())
n = int(input())
lst = [int(input()) for _ in range(n)]
dp = [[0] * (v+1) for _ in range(n+1)]
for i in range(1,n+1):
    for j in range(v+1):
        if lst[i-1] > j:
            dp[i][j] = dp[i-1][j]
        else:
            dp[i][j] = max(dp[i-1][j],dp[i-1][j-lst[i-1]] + lst[i-1])
print(v - dp[n][v])

#题解2
v = int(input())
n = int(input())
lst = [int(input()) for _ in range(n)]
dp = [0]*(v+1)
for i in range(1,n+1):
    for j in range(v,-1,-1):
    #注意此时的遍历顺序，防止破坏上一层数据
        if j >= lst[i-1]:
            dp[j] = max(dp[j],dp[j-lst[i-1]] + lst[i-1])
print(v-dp[v])