long mod most tle type opera pac read mode

地址：http://acm.split.hdu.edu.cn/showproblem.php?pid=6000

題目：

Wash

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 64000/64000 K (Java/Others)
Total Submission(s): 1250 Accepted Submission(s): 331

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

思路：

　　第一階段，正著貪心，用set或優先隊列維護，處理出每件衣服出來的時間。

　　第二階段，倒著貪心，用set或優先隊列維護，然後維護最大值。

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 #define MP make_pair
 6 #define PB push_back
 7 typedef long long LL;
 8 typedef pair<int,int> PII;
 9 const double eps=1e-8;
10 const double pi=acos(-1.0);
11 const int K=1e6+7;
12 const int mod=1e9+7;
13 
14 namespace fastIO{
15     #define BUF_SIZE 100000
16     bool IOerror=0;
17     inline char nc(){
18         static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
19         if(p1==pend){
20             p1=buf;
21             pend=buf+fread(buf,1,BUF_SIZE,stdin);
22             if(pend==p1){
23                 IOerror=1;
24                 return -1;;
25             }
26         }return *p1++;
27     }
28     inline bool blank(char ch){
29         return ch==‘ ‘||ch==‘\n‘||ch==‘\r‘||ch==‘\t‘;
30     }
31     inline bool read(int &x){
32         char ch;
33         while(blank(ch=nc()));
34         if(IOerror)return 0;
35         for(x=ch-‘0‘;(ch=nc())>=‘0‘&&ch<=‘9‘;x=x*10+ch-‘0‘);
36         return 1;
37     }
38     #undef BUF_SIZE
39 };
40 using namespace fastIO;
41 
42 struct node
43 {
44     LL id,end;
45     node(){}
46     node(LL x,LL y){id=x,end=y;}
47     bool operator < (const node &ta) const
48     {
49         return end>ta.end||(end==ta.end&&id<ta.id);
50     }
51 };
52 int n,m,l,ta[K],tb[K];
53 LL ed[K];
54 priority_queue<node>pa,pb;
55 
56 int main(void)
57 {
58     //freopen("in.acm","r",stdin);
59     int t,cs=1;read(t);
60     while(t--)
61     {
62         LL ans=0;
63         while(pa.size())pa.pop();
64         while(pb.size())pb.pop();
65         read(l),read(n),read(m);
66         for(int i=1;i<=n;i++)
67             read(ta[i]),pa.push(node(i*1LL,1LL*ta[i]));
68         for(int i=1;i<=l;i++)
69         {
70             LL tm=pa.top().end,id=pa.top().id;
71             ed[i]=tm;
72             pa.pop(),pa.push(node(id,tm+ta[id]));
73         }
74         for(int i=1;i<=m;i++)
75             read(tb[i]),pb.push(node(i*1LL,1LL*tb[i]));
76         for(int i=l;i;i--)
77         {
78             LL tm=pb.top().end,id=pb.top().id;
79             ans=max(ans,ed[i]+tm);
80             pb.pop(),pb.push(node(id,tm+tb[id]));
81         }
82         printf("Case #%d: %lld\n",cs++,ans);
83     }
84     return 0;
85 }

hdu6000 Wash ccpc-20162017-finals B Wash