1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
|
// 826. Most Profit Assigning Work
#include "leetcode.h"
/*
* you have n jobs and m workers. you are given three arrays: difficulty,
* profit, and worker where difficulty and pofit are the difficulty and profit
* of the i'th job and worker is the ability of the jth worker. every worker can
* be assigned at most one job, but one job can be completed multiple times.
* return the maximum profit we can achieve after assigning the workers to the
* jobs.
*/
int cmp(const void **a, const void **b) {
int *a1 = *(int **)a;
int *b1 = *(int **)b;
return a1[0] == b1[0] ? a1[1] - b1[1] : a1[0] - b1[0];
}
int find_profit_idx(int **set, int difficulty_size, int capability) {
int left = 0, right = difficulty_size - 1, mid;
while (left + 1 < right) {
mid = (left + right) / 2;
if (set[mid][0] <= capability)
left = mid;
else
right = mid - 1;
}
if (set[right][0] <= capability)
return right;
else if (set[left][0] <= capability)
return left;
else
return -1;
}
int maxProfitAssignment(int *difficulty, int difficultySize, int *profit,
int profitSize, int *worker, int workerSize) {
int i, j;
int **set = (int **)malloc(difficultySize * sizeof(int *));
for (int i = 0; i < difficultySize; i++) {
set[i] = malloc(2 * sizeof(int));
set[i][0] = difficulty[i];
set[i][1] = profit[i];
}
qsort(set, difficultySize, sizeof(int *), cmp);
int *dp = malloc(difficultySize * sizeof(int));
dp[0] = set[0][1];
for (int i = 1; i < difficultySize; i++)
dp[i] = set[i][1] > dp[i - 1] ? set[i][1] : dp[i - 1];
int ans = 0, idx;
for (int i = 0; i < workerSize; i++) {
idx = find_profit_idx(set, difficultySize, worker[i]);
if (idx != -1)
ans += dp[idx];
}
for (int i = 0; i < difficultySize; i++)
free(set[i]);
free(set), free(dp);
return ans;
}
int main() {
int d1[] = {2, 4, 6, 8, 10}, p1[] = {10, 20, 30, 40, 50}, w1[] = {4, 5, 6, 7};
int d2[] = {85, 47, 57}, p2[] = {24, 66, 99}, w2[] = {40, 25, 25};
printf("%d\n", maxProfitAssignment(d1, ARRAY_SIZE(d1), p1, ARRAY_SIZE(p1), w1,
ARRAY_SIZE(w1))); // expect: 100
printf("%d\n", maxProfitAssignment(d2, ARRAY_SIZE(d2), p2, ARRAY_SIZE(p2), w2,
ARRAY_SIZE(w2))); // expect: 0
}
|