ACM: 线段树 poj 3468-优快云博客

本文介绍了一种使用线段树解决区间更新与查询问题的经典算法。通过维护每个节点的累加和与增量域，可以高效地处理区间内数值的批量增加及区间和的查询。文章详细解释了解题思路，并提供了完整的C++实现代码。

A Simple Problem with Integers

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a₊₁, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

4
55
9
15

Hint

The sums may exceed the range of 32-bit integers.

题意: 一组数组, 对它有两种操作, Q A B: 访问区间[A,B]的数组累加和, C A B C: 区间[A,B]内每个

元素都加上C.

解题思路:

1. 经典的线段树题目, 只需在线段树每个节点上设置sum和add: 累加和,增量域. 题目迎刃而解.

2. 操作C时, 当满足当前区间[A,B]时, p->add = p->add + C; 修改区间的增量即可.

当不满足时, 继续往左右区间改变增量之前要将当前的区间增量加入sum,

p->sum = p->sum + (p->r - p->l + 1) * p->add;

操作Q时,当满足当前区间[A,B]时, 记录结果result = p->sum + (p->r - p->l + 1)*p->add;

当不满足时, 继续往左右区间查找之前要将当前的增量分配到左右子树区间上,

p->left->add += p->add; p->right->add += p->add; p->add = 0; //最后记得清除.

p->sum += (p->r - p->l + 1)*p->add;

3. 最后, 要注意累加的结果会超出32位整型, 并且传入C增量时要用整型64位.

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 100005

struct node
{
int l, r;
__int64 sum, add; //sum:区间和, add:区间的增量
}pt[MAX*4];

int n, m;
__int64 g[MAX];
__int64 a, b, c;
char ch[2];
__int64 result;

void buildTree(int l, int r, int pos)
{
pt[pos].l = l, pt[pos].r = r;
pt[pos].add = 0;
if(l == r)
{
pt[pos].sum = g[l];
return ;
}
int mid = (pt[pos].l+pt[pos].r)/2;
buildTree(l, mid, pos*2);
buildTree(mid+1, r, pos*2+1);
pt[pos].sum = pt[pos*2].sum+pt[pos*2+1].sum;
}

void add(int l, int r, int pos, __int64 c)
{
if(pt[pos].l == l && pt[pos].r == r)
{
pt[pos].add += c;
return ;
}

int mid = (pt[pos].l+pt[pos].r)/2;
pt[pos].sum += (r-l+1)*c;
if(r <= mid)
  add(l, r, pos*2, c);
else if(l > mid)
  add(l, r, pos*2+1, c);
else
{
  add(l, mid, pos*2, c);
  add(mid+1, r, pos*2+1, c);
}
}

void find(int l, int r, int pos)
{
if(pt[pos].l == l && pt[pos].r == r)
{
result += (pt[pos].sum + (r-l+1)*pt[pos].add);
return ;
}

int mid = (pt[pos].l+pt[pos].r)/2;
pt[pos].sum += (pt[pos].r-pt[pos].l+1)*pt[pos].add;
pt[pos*2].add += pt[pos].add;
pt[pos*2+1].add += pt[pos].add;
pt[pos].add = 0;
if(r <= mid)
  find(l, r, pos*2);
else if(l > mid)
  find(l, r, pos*2+1);
else
{
  find(l, mid, pos*2);
  find(mid+1, r, pos*2+1);
}
}

int main()
{
int i;
// freopen("input.txt","r",stdin);
while(scanf("%d %d",&n, &m) != EOF)
{
  for(i = 1; i <= n; ++i)
   scanf("%I64d",&g[i]);
  buildTree(1, n, 1);

  for(i = 1; i <= m; ++i)
  {
   scanf("%s",ch);
   if(ch[0] == 'C')
   {
    scanf("%I64d %I64d %I64d",&a, &b, &c);
    add(a, b, 1, c);
   }
   else if(ch[0] == 'Q')
   {
    scanf("%d %d",&a, &b);
    result = 0;
    find(a, b, 1);
    printf("%I64d\n",result);
   }
  }
}
return 0;
}