线段树 区间更新 访问POJ3468 A Simple Problem with Integers解题报告

A Simple Problem with Integers

Time Limit: 5000MS

Memory Limit: 131072K

Total Submissions: 78802

Accepted: 24292

Case Time Limit: 2000MS

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.

Source

POJ Monthly--2007.11.25, Yang Yi

Lazy思想 更改父亲节点后赞时不更新左右儿子。等需要访问左右儿子时再更新。避免了不必要的操作。

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#define MID(x,y) ( (x+y)>>1 )

#define L(x) ( x<<1 )

#define R(x) ( x<<1|1 )

#define MAXN 100005

 struct {

 intleft,right;

 __int64 sum;        //该区间元素的总和

 __int64 delta;      //该区间每个元素的总增量

}segTree[3*MAXN];

int num[MAXN],N,Q;

 

void lazy(int father)

{//下放过程,lazy步骤，更新左右儿子

 if (segTree[father].delta )

 {

 segTree[L(father)].sum += ( segTree[L(father)].right -segTree[L(father)].left + 1 )*segTree[father].delta;

 segTree[R(father)].sum += ( segTree[R(father)].right -segTree[R(father)].left + 1 )*segTree[father].delta;

 segTree[L(father)].delta += segTree[father].delta;

 segTree[R(father)].delta += segTree[father].delta;

 segTree[father].delta = 0;

 }

}

 

void Build(int father,int left,intright)   //口诀：赋值,判断（叶子）,递归,(回溯)更新

{

 segTree[father].left = left;segTree[father].right = right;//赋值

 if (left == right )    //判断

 {

 segTree[father].sum = num[left]; return;

 }

 intmid = MID(left,right);//递归

 Build(L(father),left,mid);

 Build(R(father),mid+1,right);

 

 segTree[father].sum = segTree[L(father)].sum +segTree[R(father)].sum;  //建完子树对根节点进行更新

}

 

void Add(int father,int left,int right,__int64numAdd)

{

 if (segTree[father].left == left && segTree[father].right == right )   //下放终止条件

 {

 segTree[father].sum += numAdd*( segTree[father].right -segTree[father].left + 1 );

 segTree[father].delta += numAdd;//Lazy关键一步，暂时存储增量而不是更新儿子。下次用到该节点的儿子时再更新儿子

 return;

 }

 

 lazy(father);    //下降一层

 

 intmid = MID(segTree[father].left,segTree[father].right);

 if (right <= mid ) Add(L(father),left,right,numAdd);

 elseif ( mid < left ) Add(R(father),left,right,numAdd);

 else

 {

 Add(L(father),left,mid,numAdd);

 Add(R(father),mid+1,right,numAdd);

 }

 //前面递归将father的左右儿子节点加了之后，回溯计算父亲节点应该加上多少

 segTree[father].sum = segTree[L(father)].sum +segTree[R(father)].sum;

}

 

__int64 query(int father,int left,intright)

{

 if (segTree[father].left == left && segTree[father].right == right )

 return segTree[father].sum;

 

 lazy(father);

 

 intmid = MID(segTree[father].left,segTree[father].right);

 if (right <= mid )

 return query(L(father),left,right);

 elseif ( mid < left )  returnquery(R(father),left,right);

 else return query(L(father),left,mid) + query(R(father),mid+1,right);

}

 

int main()

{

//freopen("output.txt","w",stdout);

 //freopen("input.txt","r",stdin);

 inti,from,to;

 __int64 ans,numAdd;

 charcmd;

 

 scanf("%d%d",&N,&Q);

 for(i=1; i<=N; i++) scanf("%d",&num[i]);

 Build(1,1,N);

 while ( Q-- )

 {

 scanf("%1s",&cmd);

  if( cmd == 'C' )

  {

  scanf("%d%d%I64d",&from,&to,&numAdd);

  Add(1,from,to,numAdd);

  }

 else

  {

  scanf("%d%d",&from,&to);

   ans= query(1,from,to);

  printf("%I64d\n",ans);

  }

 }

 return 0;

}