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// 2542. Maximum Subsequence Score
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
/*
* given two 0-indexed integer arrays 'num1' and 'nums2' of equal length 'n' and
* a positive integer 'k'. you must choose a subsequence of indices from 'nums1'
* of length 'k'. for chosen indices 'i0, i1, ..., ik-1', your score is defined
* as
* - the sum of the selected elements from 'nums1' multiplied with the minimum
* of the selected elements from 'nums2'
* - it can defined simply as '(num1[i0] + nums1[i1] + ... + nums1[ik - 1]) *
* min(nums2[i0], nums2[i1], ..., nums2[ik - 1])' return the maximum possible
* score. a subseqnce of indices of an array is a set that can be derived from
* the set '{0, 1, ..., n - 1}' by deleting no elements
*/
int cmp(const void *a, const void *b) {
int *ptr_a = *(int **)a;
int *ptr_b = *(int **)b;
if (ptr_a[1] == ptr_b[1])
return ptr_a[0] - ptr_b[0];
return ptr_a[1] - ptr_b[1];
}
int cmp1(const void *a, const void *b) { return *(int *)a - *(int *)b; }
void insert(int *heap, int k, int val, long long *sum) {
if (val <= heap[0])
return;
int f = 0;
*sum = *sum + val - heap[0];
heap[0] = val;
while (1) {
int child1 = 2 * f + 1;
int child2 = 2 * f + 2;
if (child2 < k) {
if (heap[child1] <= heap[child2]) {
if (heap[child1] < heap[f]) {
int tmp = heap[child1];
heap[child1] = heap[f];
heap[f] = tmp;
f = child1;
} else
break;
} else {
if (heap[child2] < heap[f]) {
int tmp = heap[child2];
heap[child2] = heap[f];
heap[f] = tmp;
f = child2;
} else
break;
}
} else if (child1 < k) {
if (heap[child1] < heap[f]) {
int tmp = heap[child1];
heap[child1] = heap[f];
heap[f] = tmp;
f = child1;
} else
break;
} else
break;
}
}
long long maxScore(int *nums1, int nums1_size, int *nums2, int nums2_size,
int k) {
long long ans = 0, sum = 0;
int n = nums1_size;
int **tmp = malloc(n * sizeof(int *));
int *heap = malloc(2 * k * sizeof(int));
for (int i = 0; i < n; i++) {
tmp[i] = malloc(2 * sizeof(int));
tmp[i][0] = nums1[i];
tmp[i][1] = nums2[i];
}
qsort(tmp, n, sizeof(int *), cmp);
for (int i = 0; i < k; i++) {
heap[i] = tmp[n - 1 - i][0];
sum += tmp[n - 1 - i][0];
}
ans = sum * (long long)tmp[n - k][1];
qsort(heap, k, sizeof(int), cmp1);
for (int i = n - k - 1; i >= 0; i--) {
insert(heap, k, tmp[i][0], &sum);
ans = fmax(ans, sum * tmp[i][1]);
}
return ans;
}
int main() {
int n11[] = {1, 3, 3, 2}, n21[] = {2, 1, 3, 4};
int n12[] = {4, 2, 3, 1, 1}, n22[] = {7, 5, 10, 9, 6};
printf("%lld\n", maxScore(n11, 4, n21, 4, 3)); // expect: 12
printf("%lld\n", maxScore(n12, 5, n22, 5, 1)); // expect: 30
}
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