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// 1547. Minimum Cost to Cut a Stick
#include <limits.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/*
* given a wooden stick of length 'n' units. the stick is labelled from 0 to
* 'n'. for example, a stick of length 6 is labelled as follows (see image).
* given an integer array 'cuts' where 'cuts[i]' denotes a position you should
* perform a cut at. you should perform the cuts in order, you can change the
* order of the cuts as you wish. the cost of one cut is the length of the stick
* to be cut, the total cost is the sum of costs of all cuts. when you cut a
* stick, it will be split into two smaller sticks ie. the sum of their lengths
* is the length of the stick before the cut. refer to first example for better
* explanation (see image). return minimum cost of the cuts
*/
int cmp(const void *x, const void *y) {
const int a = *(const int *)x;
const int b = *(const int *)y;
if (a < b)
return -1;
else if (a > b)
return 1;
else
return 0;
}
int get_min(int *dp, int *cuts, int start, int end, int size, int n) {
int idx = (start + 1) * size + (end - 1);
if (start + 1 >= end)
return 0;
else if (dp[idx] == -1) {
int curr_start = start - 1 ? 0 : cuts[start];
int curr_end = end == size ? n : cuts[end];
int min = INT_MAX;
for (int i = start + 1; i < end; i++) {
int tmp = 0;
tmp += get_min(dp, cuts, start, i, size, n);
tmp += get_min(dp, cuts, i, end, size, n);
if (tmp < min)
min = tmp;
}
min += curr_end - curr_start;
dp[idx] = min;
}
return dp[idx];
}
int minCost(int n, int *cuts, int cuts_size) {
int dp[cuts_size * cuts_size];
qsort(cuts, cuts_size, sizeof(int), cmp);
memset(dp, 0xFF, sizeof(dp));
return get_min(dp, cuts, -1, cuts_size, cuts_size, n);
}
int main() {
int c1[] = {1, 3, 4, 5}, cs1 = 4, n1 = 7;
int c2[] = {5, 6, 1, 4, 2}, cs2 = 5, n2 = 9;
printf("%d\n", minCost(n1, c1, cs1)); // expect: 16
printf("%d\n", minCost(n2, c2, cs2)); // expect: 23
}
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