Luogu2751 [USACO Training4.2]工序安排Job Processing

你怎么搞他时间都是一样的，考虑贪心

计算第一轮的时间就直接拿堆维护每个机器的结束时间，
每次取最早的，更新每个物品的答案

考虑第二轮，每次取上轮结束时间最晚的放到结束时间最早的机器里
这样一定是最优的

由于 a 类机器和 b 类机器是分开工作的，所以就直接这样贪心就好了

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <cstdio>
#include <queue>
using namespace std;

const int MAXN = 1005, MAXM = 35;

struct Node{
	int tim, len;
	bool operator < (const Node& b) const {
		return tim > b.tim;
	}
}a[MAXM], b[MAXM];
int n, ma, mb, maxans;
int fa[MAXN], fb[MAXN];
priority_queue<Node> q;

int main() {
	scanf("%d%d%d", &n, &ma, &mb);
	for (int i = 1; i <= ma; ++i) {
		scanf("%d", &a[i].len);
		a[i].tim = a[i].len;
		q.push(a[i]);
	}
	for (int i = 1; i <= mb; ++i) {
		scanf("%d", &b[i].len);
		b[i].tim = b[i].len;
	}
	for (int i = 1; i <= n; ++i) {
		Node tmp = q.top();
		q.pop();
		fa[i] = tmp.tim;
		tmp.tim += tmp.len;
		q.push(tmp);
	}
	printf("%d ", fa[n]);
	while (q.size()) q.pop();
	for (int i = 1; i <= mb; ++i)
		q.push(b[i]);
	for (int i = n; i >= 1; --i) {
		Node tmp = q.top();
		q.pop();
		maxans = max(maxans, fb[i] = fa[i] + tmp.tim);
		tmp.tim += tmp.len;
		q.push(tmp);
	}
	printf("%d\n", maxans);
	return 0;
}

原文地址：https://www.cnblogs.com/xcysblog/p/9800815.html