加工生产调度

题目描述

思路

代码

#include <cstdio>

int n, na, nb, ans;
struct Node {
	int a, b, c;
} aa[1005], bb[1005], tmp;
inline int read() {
	int s = 0, f = 1;
	char ch = getchar();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
	return s * f;
}
void qsortA(int L, int R, Node * arr) {
	int l = L, r = R;
	int mid = arr[l + r >> 1].a;
	while (l < r) {
		while (arr[l].a < mid) l++;
		while (arr[r].a > mid) r--;
		if (l <= r) {
			tmp = arr[l];
			arr[l] = arr[r];
			arr[r] = tmp;
			l++, r--;
		}
	}
	if (l < R) qsortA(l, R, arr);
	if (L < r) qsortA(L, r, arr);
}
void qsortB(int L, int R, Node * arr) {
	int l = L, r = R;
	int mid = arr[l + r >> 1].b;
	while (l < r) {
		while (arr[l].b > mid) l++;
		while (arr[r].b < mid) r--;
		if (l <= r) {
			tmp = arr[l];
			arr[l] = arr[r];
			arr[r] = tmp;
			l++, r--;
		}
	}
	if (l < R) qsortB(l, R, arr);
	if (L < r) qsortB(L, r, arr);
}
int main() {
	n = read();
	for (int i = 1; i <= n; ++i) aa[i].a = read();
	for (int i = 1; i <= n; ++i) aa[i].b = read();
	for (int i = 1; i <= n; ++i) {
		if (aa[i].a < aa[i].b) {
			aa[++na] = aa[i], aa[na].c = i;
		} else {
			bb[++nb] = aa[i], bb[nb].c = i;
		}
	}
	qsortA(1, na, aa); qsortB(1, nb, bb);
	for (int j = 1; j <= nb; ++j) aa[++na] = bb[j];
	for (int i = 1, a = 0, b = 0; i <= na; ++i) {
		a += aa[i].a;
		if (ans < a) {
			ans = a + aa[i].b;
		} else {
			ans += aa[i].b;
		}
	}
	printf("%d\n", ans);
	for (int i = 1; i <= na; ++i) {
		printf("%d", aa[i].c);
		if (i != na) putchar(' ');
	}
	return 0;
}