hdu 4020のとても良いsetと構造体の結合はとてもかっこいい1つの問題を運用します

Ads Proposal
Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others) Total Submission(s): 1537    Accepted Submission(s): 534
Problem Description
There are N customers set their M different advertisements on Baidu. Each advertisement is owned by single customer. The ads system of Baidu records the number of clicks of each advertisement this year. The proposal system wants to analysis the advertisements about the relativity of the description length and the number of clicks of the advertisements. During the analysis, it is very important to do such a query to ask the total length of the advertisements with top k clicking times for each customer. You may assume the number of clicks of all advertisements are distinct.
Your task is to help Baidu to design this little toolkit.
 
Input
The input consist multiple test cases. The number of test cases is given in the first line of the input.
  For each test case, the first line contains three integers N, M and Q, denoting the number customer, the number of advertisement instances and the number of queries. (N <= 100000, M <= 500000, Q <= 100000)
  Then M lines follow, each line contains three numbers, U, C and L, indicating the owner of this advertisement, the clicks for this advertisement and the length. (1 <= U <= N, 0 <= C, L <= 1000000000)
  Finally Q lines come. Each line contains only one integer k, representing the query for top k clicking advertisements for each customer.
 
Output
For each test case, output Q lines, each line contains only one integer, denoting the sum of total length of the top k number of clicks for each customer.
 
Sample Input

   
   
   
   

    
    
    
    2
2 4 3
1 12 13
2 23 41
1 21 46
1 22 31
1
2
3
6 15 3
5 2677139 731358928
2 347112028 239095183
6 27407970 85994789
6 767687908 734935764
6 255454855 110193353
3 39860954 813158671
5 617524049 55413590
3 338773814 7907652
6 810348880 736644178
2 777664288 63811422
6 590330120 616490361
5 552407488 136492190
1 416295130 448298060
5 811513162 232437061
4 43273262 874901209
4
9
13
   
   
   
   

 
Sample Output

   
   
   
   

    
    
    
    Case #1:
72
118
131
Case #2:
5801137622
5887132411
5887132411
   
   
   
   

 
Source
The 36th ACM/ICPC Asia Regional Shanghai Site —— Warmup
 
Recommend
lcy
題意説明:n個の顧客、m個の広告があって、各顧客は複数の広告を持つことができて、各広告は1つのクリック量と長さがあって、今各顧客の前のk個のアクセス量の長さの総和の総和を要求します
考え方:
kごとに求めないで  次に加算    プログラムの中で表を打ってすべてを探し出すべきです    ダイレクト対応出力

#include<stdio.h>
#include<string.h>
#include<set>
using namespace std;
#define ll __int64
struct haha
{
	ll c;
	ll l;
	bool operator<(const haha &x) const  //  set                             
	{
		return c>x.c;//        
	}
}a;
set<haha>ss[100000+10];
set<haha>::iterator it;
int getval()//     
{     
    int ret(0);     
    char c;     
    while((c=getchar())==' '||c=='
'||c=='\r');     
        ret=c-'0';     
    while((c=getchar())!=' '&&c!='
'&&c!='\r')                 
            ret=ret*10+c-'0';     
        return ret;
}
ll ans[500100];
int main()
{
	int i,cas,n,m,q,c_num=0,j;
//	scanf("%d",&cas);
	cas=getval();
	while(cas--)
	{
	//	scanf("%d %d %d",&n,&m,&q);//                                      
		n=getval();
		m=getval();
		q=getval();
		for(i=0;i<=n;i++) ss[i].clear();// ！！！！！     =                  
		for(i=0;i<m;i++)
		{
			int  user;
			user=getval();
			a.c=(ll)getval();
			a.l=(ll)getval();
		//	scanf("%d %I64d %I64d",&user,&a.c,&a.l);
			ss[user].insert(a);
		}
		memset(ans,0,sizeof(ans));
		for(i=1;i<=n;i++)
		{
			  int cnt=1;
			  for(it=ss[i].begin();it!=ss[i].end();it++)
			  {
                         ans[cnt]+=(*it).l;
						 cnt++;
			  }
		}	
        for(i=1;i<=m;i++) 
		{
            ans[i]+=ans[i-1];
        }
		printf("Case #%d:
",++c_num);
		for(i=0;i<q;i++)
		{
			int k;
			k=getval();
			if(k>m)  k=m; 
			printf("%I64d
",ans[k]);
		}
	}
	return 0;
}