hdu 4288 Coder 线段树 区间合并
Coder
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4749 Accepted Submission(s): 1808
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
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 5 ), 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 9.
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).
In each test case, the first line contains one integer N ( 1 <= N <= 10 5 ), 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 9.
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.
Source
2012 ACM/ICPC Asia Regional Chengdu Online
Recommend
liuyiding | We have carefully selected several similar problems for you: 4297 4289 4290 4291 4292
1.通过最近几个题发现，线段树可以解决在连续区间内一共存在多少个值的问题（某些点存在，某一点不存在）
查询的时候既可以查询出某一段区间的sum之和，又可以在一段连续，却不一定每个点都存在的区间上，去统计找到k个值的位置。
线段树的灵活运用在于update的顺序性。
2.
该题目区间合并的方式：
void merge(int ind)
{
int c=tree[ind*2].sum;
for(int i=0;i<5;i++) tree[ind].mod[i]=tree[ind*2].mod[i];
for(int i=0;i<5;i++) tree[ind].mod[(i+c)%5]+=tree[ind*2+1].mod[i];
}
{
int c=tree[ind*2].sum;
for(int i=0;i<5;i++) tree[ind].mod[i]=tree[ind*2].mod[i];
for(int i=0;i<5;i++) tree[ind].mod[(i+c)%5]+=tree[ind*2+1].mod[i];
}
因为 题目所求是排序后序数 %5=3的元素之和。
3.
STL技能：vector删重:
ve.erase(unique(ve.begin(),ve.end()),ve.end());
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
2017.7.12:
难点在于合并公式的推导：
#include
#include
#include
#include
#include
#include
#include