Coder（線段樹+多棵線段樹）

Coder

Time Limit : 20000/10000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 12   Accepted Submission(s) : 7

Problem Description 　　In mathematics and computer science, an algorithm describes a set of procedures or instructions that define a procedure. The term has become increasing popular since the advent of cheap and reliable computers. Many companies now employ a single coder to write an algorithm that will replace many other employees. An added benefit to the employer is that the coder will also become redundant once their work is done.  1

　　You are now the signle coder, and have been assigned a new task writing code, since your boss would like to replace many other employees (and you when you become redundant once your task is complete).

Your code should be able to complete a task to replace these employees who do nothing all day but eating: make the digest sum.

　　By saying “digest sum” we study some properties of data. For the sake of simplicity, our data is a set of integers. Your code should give response to following operations:

　　1. add x – add the element x to the set;

　　2. del x – remove the element x from the set;

　　3. sum – find the digest sum of the set. The digest sum should be understood by

　　where the set S is written as {a 1, a 2, ... , a k} satisfying a 1 < a 2 < a 3 < ... < a k 

　　Can you complete this task (and be then fired)?

------------------------------------------------------------------------------

1 See http://uncyclopedia.wikia.com/wiki/Algorithm  

Input 　　There’re several test cases. 　　In each test case, the first line contains one integer N ( 1 <= N <= 10[sup]5[/sup] ), the number of operations to process. 　　Then following is n lines, each one containing one of three operations: “add x” or “del x” or “sum”. 　　You may assume that 1 <= x <= 10[sup]9[/sup]. 　　Please see the sample for detailed format. 　　For any “add x” it is guaranteed that x is not currently in the set just before this operation. 　　For any “del x” it is guaranteed that x must currently be in the set just before this operation. 　　Please process until EOF (End Of File).  

Output 　　For each operation “sum” please print one line containing exactly one integer denoting the digest sum of the current set. Print 0 if the set is empty.  

Sample Input

9
add 1
add 2
add 3
add 4
add 5
sum
add 6
del 3
sum
6
add 1
add 3
add 5
add 7
add 9
sum

         

Sample Output

3
4
5
[hint]C++ maybe run faster than G++ in this problem.[/hint]
         

Source 2012 ACM/ICPC Asia Regional Chengdu Online  

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題意：

維護一個有序數列{An}，有三種操作：

1、添加一個元素。

2、删除一個元素。

3、求數列中下标%5 = 3的值的和。

由于有删除和添加操作，是以離線離散操作，節點中cnt存儲區間中有幾個數，sum存儲偏移和

代碼：

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<algorithm>  
using namespace std;  
typedef long long ll;
#define maxn 100002
#define lson l , m , rt << 1    
#define rson m + 1 , r , rt << 1 | 1 
#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)
   
ll sum[maxn<<2][6]; //存儲偏移和
 
int cnt[maxn << 2];   //存儲區間中有幾個數

char op[maxn][20];  

int a[maxn];  
  
int X[maxn];  
  
void PushUp(int i) 
{  
    cnt[i] = cnt[L(i)] + cnt[R(i)];  
      
    int t=cnt[L(i)];  
    for(int j = 0; j < 5; ++j)  
    {  
        sum[i][j] = sum[L(i)][j];  
    }  
    for(int k=0;k < 5; ++k)  
    {  
        sum[i][(k + t) % 5] += sum[R(i)][k];  
    }  
}  
 
void Build(int l, int r, int rt)    
{ 
    cnt[rt] = 0;
    for(int i = 0; i < 5; ++i)   
    sum[rt][i] = 0;
    if( l == r ) 
    return;  
    int m= (l+r)>>1;      
    Build(lson);    
    Build(rson);     
}   
  
void Updata(int p, int op, int l, int r, int rt)    
{     
    if(l==r)    
    {    
        cnt[rt]=op;   
        sum[rt][1]=op*X[l-1];   
        return;    
    }    
    int m = ( l + r ) >> 1;    
    if(p <= m)    
        Updata(p, op, lson);    
    else    
        Updata(p, op, rson);    
    
    PushUp(rt);    
}   
  
int main()  
{  
    int n;  
    while(scanf("%d", &n) != EOF)  
    {  
        int nn = 0;  
        for(int i = 0; i < n; ++i)  
        {  
            scanf("%s", &op[i]);  
              
            if(op[i][0] != 's')  
            {  
                scanf("%d", &a[i]);  
                if(op[i][0] == 'a')  
                {  
                    X[nn++] = a[i];  
                }  
            }  
        }  
          
        sort(X,X+nn);//unique前必須sor
        nn = unique(X, X + nn) - X;//去重并得到總數
        Build(1, nn, 1);
        for(int i = 0; i < n; ++i)  
        {  
            int pos = upper_bound(X, X+nn, a[i]) - X;
            if(op[i][0] == 'a')  
            {  
                Updata(pos, 1, 1, nn, 1);  
            }  
            else if(op[i][0] == 'd') 
            {  
                Updata(pos, 0, 1, nn, 1);  
            }  
            else printf("%lld\n",sum[1][3]);  
        }  
    }  
    return 0;  
}