[贪心 优先队列] 两步骤完成多项工作总用时最短 HDU6000

Wash

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 64000/64000 K (Java/Others)
Total Submission(s): 2492    Accepted Submission(s): 672

 

Problem Description

Mr.Panda is about to engage in his favourite activity doing laundry! He’s brought L indistinguishable loads of laundry to his local laundromat, which has N washing machines and M dryers.The ith washing machine takes Wi minutes to wash one load of laundry, and the ith dryer takes Di minutes to dry a load of laundry.
At any point in time, each machine may only be processing at most one load of laundry.
As one might expect, Panda wants to wash and then dry each of his L loads of laundry. Each load of laundry will go through the following steps in order:
1. A non-negative amount of time after Panda arrives at the laundromat, Panda places the load in an unoccupied washing machine i.
2. Wi minutes later, he removes the load from the washing machine, placing it in a temporary holding basket (which has unlimited space)
3. A non-negative amount of time later, he places the load in an unoccupied dryer j 
4. Dj minutes later, he removes the load from the dryer Panda can instantaneously add laundry to or remove laundry from a machine. Help Panda minimize the amount of time (in minutes after he arrives at the laundromat) after which he can be done washing and drying all L loads of laundry!

 

 

Input

The first line of the input gives the number of test cases, T.
T test cases follow. Each test case consists of three lines. The first line contains three integer L, N, and M.
The second line contains N integers W1,W2,...,WN representing the wash time of each wash machine.
The third line contains M integers D1,D2,...,DM representing the dry time of each dryer.

 

 

Output

For each test case, output one line containing “Case #x: y”, where x is the test case number (starting from 1) and y is the minimum time it will take Panda to finish his laundry.

limits

∙1≤T≤100.
∙1≤L≤106.
∙1≤N,M≤105.
∙1≤Wi,Di≤109.

 

 

Sample Input

2

1 1 1

1200

34

2 3 2

100 10 1

10 10

 

 

Sample Output

Case #1: 1234

Case #2: 12

 

 

Source

2016 CCPC-Final

 

 

使总时间最短
洗衣完成较早者对应烘干所需时间最小的L项中较迟者
时限卡得很紧

#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int ml = 1e6 + 10;

ll A[ml];
struct node
{
	int id;
	ll time;
} a[ml], b[ml];
bool operator <(const node& a, const node& b)
{
	return a.time > b.time;
}
priority_queue<node> q1, q2;

int main()
{
	int T;
	scanf("%d", &T);
	for (int cas = 1; cas <= T; cas++)
	{
		while (!q1.empty())
			q1.pop();
		while (!q2.empty())
			q2.pop();
	
		int l, n, m;
		scanf("%d %d %d", &l, &n, &m);
		for (int i = 1; i <= n; i++)
		{
			a[i].id = i;
			scanf("%lld", &a[i].time);
			q1.push(a[i]);
		}
		for (int i = 1; i <= m; i++)
		{
			b[i].id = i;
			scanf("%lld", &b[i].time);
			q2.push(b[i]);
		}
		
		// 洗衣完成时间从小到大
		for (int i = 0; i < l; i++)
		{
			node t = q1.top();
			A[i] = t.time;
			q1.pop();
			t.time += a[t.id].time;
			q1.push(t);
		}
		
		ll ans = 0;
		// 烘干完成时间与 A[i] 反着对应
		for (int i = l - 1; i >= 0; i--)
		{
			node t = q2.top();
			ans = max(ans, t.time+ A[i]);
			q2.pop();
			t.time += b[t.id].time;
			q2.push(t);
		}
		printf("Case #%d: %lld\n", cas, ans);
	}
	return 0;
}