LintCode 1161: goods transfer (DP超难题)

1161. goods transfer

There are n backpacks, each of which can carry a[i] goods, and the maximum capacity of each backpack is b[i]. If it takes 1 second to transfer one piece of goods, how many backpacks can hold all the goods at least, and how much time does it take to transfer?

Example

Example

Input
4
[3, 3, 4, 3]
[4, 7, 6, 5]
Output 2 6 
Explanation:
We can put the goods in the first bag all moved to the second backpack, take time for 3 seconds, at this point the second of the residual capacity of backpack is 1, and then put the goods in the fourth incubator transfer a into the second bag, transfer the last two to the third in the backpack. Take time for 3 seconds, so the minimum heat preservation box number is 2, at least spend time for 6 seconds.

Notice

• 1<=n<=100
• 1<=ai<=1000
• 1<=bi<=1000
• ai<=bi

Input test data (one parameter per line)How to understand a testcase?

解法1：
这题非常难，我是照抄的网上的答案然后学习。

首先这题求2个东西，
1）how many backpacks can hold all the goods at least,
2）how much time does it take to transfer?
1)是求最少包数；2)是在1)的基础上最少时间，即最少转移的食物piece数。

首先1)可以用贪婪法来解决。比如说a = {3,3,4,3}, b = {4,7,6,5}，因为sum_a = 13，而b里面找尽量大的元素，使得其和加起来大于sum_a即可。
但2)不能用贪婪法，比如说1)的最少包数决定后，比如说X个包，然后2)是不是就找最大的X个包就可以了呢？答案是否定的。比如说a={4,1}, b={5,6}。可知用sum_a=5，可知用1个包就可以了，但取6是错的，因为要转移的食物数为4。实际上应该取5，此时要转移的食物数为1。

最后是只能用DP。注意要同时保存2个数组。dp和weight。
dp[i][j]表示前i个箱子，挑出最少的箱子个数后构成j的容量时，那个最少的箱子总数。注意dp[i][j]可以优化成1维数组dp[j]。
weight[j]表示上面的最优方案里这些箱子里的食物之和最大是多少。

转移方程怎么写呢？

                int res = max(j - b[i], 0);
                dp[j] = min(dp[j],dp[max(0, j-b[i])]+1)
即取b[i]后，如果dp个数更少，则更新dp[j]和weight[j]。如果dp个数一样，则weight[j]取大的情况。
注意：
1) 转移方程应以b[]为主，因为b[i]不能拆分，而不能以a[]为主，因为a[i]可以拆分。
2) dp越小越优，当dp相等时weight越大越优。
3) 一个难点是j循环内j<=b[i]时的处理。注意这里跟01背包问题不一样，01背包问题的j的下限就是A[i]。这里还要处理0<j<=b[j]的情况呢？我的理解是01背包问题里面，A[i]是物品的大小，k<A[i]时，A[i]只可能不取，那么dp值就是不取A[i]的情况，不变即可。
见https://blog.youkuaiyun.com/roufoo/article/details/102996040
这里b[j]是包的大小，即使j<=b[i]，还是可以取这个包。

class Solution {
public:
    /**
     * @param n: the number of backpacks
     * @param a: the number of goods each backpack carries
     * @param b:  the maximum capacity of each backpack
     * @return: given n, ai and bi,return the minimum number of backpacks and the minimum time
     */
    vector<int> goodsTransfer(int n, vector<int> &a, vector<int> &b) {
        int const INF = 0x3f3f3f3f;
        int k = INF, max_weight = 0;
        int sum_good = 0, sum_cap = 0;
        for (int i = 0; i < a.size(); i++) {
            sum_good += a[i];
            sum_cap += b[i];
        }
        
        //dp[i][j]表示前i个箱子，挑出最少的箱子个数后构成j的容量时，那个最少的箱子总数
        //weight[j]表示上面的最优方案里这些箱子里的食物之和最大是多少。
        vector<int> dp(sum_cap + 1, INF);
        vector<int> weight(sum_cap + 1, 0);
        dp[0] = 0;
        for (int i = 0; i < n; i++) {
            for (int j = sum_cap; j > 0; j--) {
                int res = max(j - b[i], 0);
                if (dp[j] < dp[res] + 1) {
                    continue;
                } else if (dp[j] > dp[res] + 1) {
                    dp[j] = dp[res] + 1;
                    weight[j] = weight[res] + a[i];
                } else {
                    weight[j] = max(weight[j], weight[res] + a[i]);
                }
            }
        }
    
    
       //dp[i]越小越优, 如果dp[i]结果一致，则weight[i]越大越好。
        for (int i = sum_good; i <= sum_cap; i++) {
            if (dp[i] < k) {
                k = dp[i];
                max_weight = weight[i];
            } else if (dp[i] == k) {
                max_weight = max(max_weight, weight[i]);
            }
        }
    
        return vector<int>{k, sum_good - max_weight};
    }
};