We have jobs: difficulty[i] is the difficulty of the ith job, and profit[i] is the profit of the ith job.

Now we have some workers. worker[i] is the ability of the ith worker, which means that this worker can only complete a job with difficulty at most worker[i].

Every worker can be assigned at most one job, but one job can be completed multiple times.

For example, if 3 people attempt the same job that pays 1,thenthetotalprofitwillbe3. If a worker cannot complete any job, his profit is $0.

What is the most profit we can make?

Example 1:

Input: difficulty = [2,4,6,8,10], profit = [10,20,30,40,50], worker = [4,5,6,7]
Output: 100 
Explanation: Workers are assigned jobs of difficulty [4,4,6,6] and they get profit of [20,20,30,30] seperately.
 

Notes:

• 1 <= difficulty.length = profit.length <= 10000
• 1 <= worker.length <= 10000
• difficulty[i], profit[i], worker[i] are in range [1, 10^5]

Solution 1

// by   JOHNKRAM
class Solution {
public:
    int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit, vector
 
<int>& worker) {
        int a[100005],n=difficulty.size(),m=worker.size(),ans=0,i;
        memset(a,0,sizeof(a));
        for(i=0;i<n;i++)
            a[difficulty[i]]=max(a[difficulty[i]],profit[i]);  // 每種難度只保留最大利潤值
        for(i=1;i<=100000;i++)
            a[i]=max(a[i],a[i-1]);  // 若低難度有更大利潤，則更新高難度的利潤
        for(i=0;i<m;i++)
            ans+=a[worker[i]];  // 求和即得結果
        return ans;
    }
};