标签：pop class stream mes max operator training 更新 amp
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
代码：
#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;
}
Luogu2751 [USACO Training4.2]工序安排Job Processing
标签：pop class stream mes max operator training 更新 amp
原文地址：https://www.cnblogs.com/xcysblog/p/9800815.html
