[LeetCode] 1090. Largest Values From Labels

Description

We have a set of items: the i-th item has value values[i] and label labels[i].

Then, we choose a subset S of these items, such that:

|S| <= num_wanted
For every label L, the number of items in S with label L is <= use_limit.

Return the largest possible sum of the subset S.

Example 1:
Input: values = [5,4,3,2,1], labels = [1,1,2,2,3], num_wanted = 3, use_limit = 1
Output: 9
Explanation: The subset chosen is the first, third, and fifth item.

Example 2:
Input: values = [5,4,3,2,1], labels = [1,3,3,3,2], num_wanted = 3, use_limit = 2
Output: 12
Explanation: The subset chosen is the first, second, and third item.

Example 3:
Input: values = [9,8,8,7,6], labels = [0,0,0,1,1], num_wanted = 3, use_limit = 1
Output: 16
Explanation: The subset chosen is the first and fourth item.

Example 4:
Input: values = [9,8,8,7,6], labels = [0,0,0,1,1], num_wanted = 3, use_limit = 2
Output: 24
Explanation: The subset chosen is the first, second, and fourth item.

Note:

1 <= values.length == labels.length <= 20000
0 <= values[i], labels[i] <= 20000
1 <= num_wanted, use_limit <= values.length

思路

题意：选取一个子集，子集中元素的总个数不大于 num_wanted，对于 label 值相同的元素，选取的个数不能大于 use_limit（如use_limit = 2，label值为 1 的元素有4个，最多选两个），这样的子集要求其 value 总和最大

题解：贪心，按照元素的value值排序，数值大的先选

static const auto io_sync_off = []()
{
    // turn off sync
    std::ios::sync_with_stdio(false);
    // untie in/out streams
    std::cin.tie(nullptr);
    return nullptr;
}();

struct Node {
    int value, label;

    bool operator < (const Node &node)const{
        return value > node.value;
    }
};

class Solution {
public:

    int largestValsFromLabels(vector<int> &values, vector<int> &labels, int num_wanted, int use_limit) {
        int size = values.size();
        Node node[size + 5];
        int sum = 0;
        map<int, int>mp;
        map<int, int>::iterator it;
        for (int i = 0; i < size; i++) {
            node[i].value = values[i];
            node[i].label = labels[i];
        }
        sort(node, node + size);
        for (int i = 0; i < size && num_wanted; i++) {
            if (mp.find(node[i].label) != mp.end()) {
                if (mp[node[i].label] < use_limit) {
                    sum += node[i].value;
                    mp[node[i].label]++;
                    num_wanted--;
                }
            } else {
                mp[node[i].label] = 1;
                sum += node[i].value;
                num_wanted--;
            }
        }
        return sum;
    }
};

来源：oschina

链接：https://my.oschina.net/u/4337909/blog/3496750

标签

subset