批量作业调度

最笨的方法实现批量作业调度问题。

计算出所有作业的左右排列的耗时

#include <stdio.h>
#include <stdlib.h>

int flowadd (int *arr, int *arr1, int *arr2, int len) {
	int i, dlte = 0 - arr2[arr[0]];
	int sum = arr1[arr[0]] + arr2[arr[0]];
	for (i = 1; i < len; ++i) {
		dlte += arr1[arr[i]];
		sum += (dlte > 0 ? dlte : 0);
		sum += arr2[arr[i]];
		dlte -= arr2[arr[i]];
	}
	return sum;
}

void printArr (int *arr, int len, int val) {
	int i;
	printf ("val = %02d ;  arr : ", val);
	for (i = 0; i < len; ++i) {
		printf ("%d ", arr[i]);
	}
	printf ("\n");
}


void flowshop (int *arr, int *arr1, int *arr2, int len, int pos) {
	int i;
	if (pos == len) {
		int tmp = flowadd (arr, arr1, arr2, len);
		printArr (arr, len, tmp);
		return;
	}
	for (i = pos; i < len; ++i) {
		int tmp = arr[pos];
		arr[pos] = arr[i];
		arr[i] = tmp;
		flowshop (arr, arr1, arr2, len, pos + 1);
		arr[i] = arr[pos];
		arr[pos] = tmp;
	}
}

int main (void) {
	int arr1[] = {2,3,2,3};
	int arr2[] = {1,1,3,5};
	int arr[] = {0,1,2,3};
	flowshop (arr, arr1, arr2, sizeof(arr1)/sizeof(int), 0);
	return 0;
}