HDU4288Coder(线段树+离线查询)
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4288
Coder
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5610 Accepted Submission(s): 2120
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.
题目大意：
维护一个有序数列{An}，有三种操作：
1：add x 就是把x插进去
2：delete x 就是把x删除
3：sum 就是求下标%5=3的元素的和。
题目要求插入和删除后都要保证数列有序大佬的分析：http://blog.csdn.net/dgq8211/article/details/7999179
由于线段树中不支持添加、删除操作，所以题解写的是用离线做法。
我们来看它是如何解决添加、删除的问题的。
首先将所有出现过的数记录下来，然后排序去重，最后根据去重结果建树，然后每个操作数都会对应线段树中的一个点。
遇到添加、删除操作的时候，只要把那个节点的值改变，然后将它对下标的影响处理好就可以。
那么如何处理这些操作对下标的影响呢？
现在我们考虑一个父区间，假设它的左右子区间已经更新完毕。
显然，左区间中下标%5的情况与父区间中下标%5的情况完全相同；
可是，右区间中却不一定相同，因为右区间中的下标是以 mid 当作 1 开始的。
那么，只要我们知道左区间中有效元素的个数 cnt，我们就能知道右区间中的下标 i 在父区间中对应的下标为 i+cnt。
所以，虽然我们最终要的只是总区间中下标%5 = 3的和。但是在更新时我们需要知道右区间%5的所有情况。
于是我们要在线段树的每个节点开一个 sum[5] 和一个 cnt，分别记录这个节点中下标%5的5种情况的和与有效元素的个数。
而查询时，直接访问总区间的 sum[3] 即可。
如此，题目便可解了。复杂度O(M logN logN)。
我的一句话题解：
先读入所有的数据，离散化后建立线段树。在每个结构体中设立一个cnt变量用于维护这段区间内数的个数。这道题的难点在于%5=3这个条件。通过左子区间内元素的个数，就可以知道，在右子区间内，元素的相对位置。
收获：
主要在于对取余确定位置这一技巧的掌握。这是一个非常有用的知识点！
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int maxn = 100005;
char str[maxn][10];
int num[maxn], s[maxn];
int len, flag;
struct Node
{
int l, r, cnt;
ll sum[5];//sum[i]保存%5的各组和
}segTree[maxn<<2];
int ser(int k)
{
int l = 0, r = len - 1, mid;
while(l <= r)
{
mid = (l+r) >> 1;
if(s[mid] < k)
l = mid + 1;
else if(s[mid] > k)
r = mid - 1;
else
return mid;
}
}
void build(int l, int r, int rt)
{
segTree[rt].l = l;
segTree[rt].r = r;
segTree[rt].cnt = 0;
memset(segTree[rt].sum, 0, sizeof(segTree[rt].sum));
if(l == r)
return;
int mid = (l + r) >> 1;
build(l, mid, rt<<1);
build(mid+1, r, rt<<1|1);
}
void pushup(int rt)
{
for(int i = 0; i < 5; i++)
segTree[rt].sum[i] = segTree[rt<<1].sum[i] + segTree[rt<<1|1].sum[((i-segTree[rt<<1].cnt)%5 + 5) % 5];
//这个区间的和是左右子树同样余数的相加和，右子树要用序号减去左子树的个数对5取模后为了保持正数所以要+5再取模，所以是(i-segTree[x*2].cnt)%5+5
}
void update(int rt, int pos, int val)
{
segTree[rt].cnt += 2*flag - 1;//区间内有几个数字
if(segTree[rt].l == segTree[rt].r)
{
segTree[rt].sum[0] = flag * val;//删除则此位清0，add则进行更新
return ;
}
int mid = (segTree[rt].l + segTree[rt].r) >> 1;
if(pos <= mid)
update(rt<<1, pos, val);
else
update(rt<<1|1, pos, val);
pushup(rt);
}
int main()
{
int n, pos;
while(scanf("%d", &n) != EOF)
{
len = 0;
for(int i = 0; i < n; i++)
{
scanf("%s", str[i]);
if(str[i][0] != 's')
{
scanf("%d", &num[i]);
s[len++] = num[i];
}
}
sort(s, s+len);
len = unique(s, s+len) - s;//去重
if(!len)
memset(segTree[1].sum, 0, sizeof(segTree[1].sum));
else
build(1, len, 1);
for(int i = 0; i < n; i++)
{
if(str[i][0] == 'a')
{
flag = 1;
pos = ser(num[i]);
update(1, pos, num[i]);
}
else if(str[i][0] == 'd')
{
flag = 0;
pos = ser(num[i]);
update(1, pos, num[i]);
}
else
printf("%I64d\n", segTree[1].sum[2]);
}
}
return 0;
}