POJ 3468 线段树区间求和

线段树区间更新与查询

最新推荐文章于 2019-01-21 15:13:26 发布

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最新推荐文章于 2019-01-21 15:13:26 发布

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原文链接：http://www.cnblogs.com/zero-zz/p/4811331.html

本文介绍了一种使用线段树进行区间更新与查询的方法，通过存储节点和的增量来提高效率。具体实现了线段树的构建、插入、添加操作及区间和的查询，并附带了一个整数操作问题的样例程序。

线段树区间求和树节点不能只存和，只存和，会导致每次加数的时候都要更新到叶子节点，速度太慢(O(nlogn))。所以我们要存两个量，一个是原来的和nSum，一个是累加的增量Inc。

在增加时，如果要加的区间正好覆盖一个节点，则增加其节点的Inc值，不再往下走，否则要更新nSum(加上本次增量),再将增量往下传，这样更新的复杂度就是O(log(n))。

在查询时，如果待查区间不是正好覆盖一个节点，就将节点的Inc往下带，然后将Inc清0，接下来再往下查询。 Inc往下带的过程也是区间分解的过程，复杂度也是O(log(n))。

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 79334		Accepted: 24455
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

  1 #include <iostream>
  2 #include<cstdio>
  3 using namespace std;
  4 struct CNode
  5 {
  6     int L ,R;
  7     CNode * pLeft, * pRight;
  8     long long nSum; //原来的和
  9     long long Inc; //增量c的累加
 10 };
 11 CNode Tree[200010]; // 2倍叶子节点数目就够
 12 int nCount = 0;
 13 int Mid( CNode * pRoot)
 14 {
 15     return (pRoot->L + pRoot->R)/2;
 16 }
 17 void BuildTree(CNode * pRoot,int L, int R)
 18 {
 19     pRoot->L = L;
 20     pRoot->R = R;
 21     pRoot->nSum = 0;
 22     pRoot->Inc = 0;
 23     if( L == R)
 24         return;
 25     nCount ++;
 26     pRoot->pLeft = Tree + nCount;
 27     nCount ++;
 28     pRoot->pRight = Tree + nCount;
 29     BuildTree(pRoot->pLeft,L,(L+R)/2);
 30     BuildTree(pRoot->pRight,(L+R)/2+1,R);
 31 }
 32 void Insert( CNode * pRoot,int i, int v)
 33 {
 34     if( pRoot->L == i && pRoot->R == i)
 35     {
 36         pRoot->nSum = v;
 37         return ;
 38     }
 39     pRoot->nSum += v;
 40     if( i <= Mid(pRoot))
 41         Insert(pRoot->pLeft,i,v);
 42     else
 43         Insert(pRoot->pRight,i,v);
 44 }
 45 void Add( CNode * pRoot, int a, int b, long long c)
 46 {
 47     if( pRoot->L == a && pRoot->R == b)
 48     {
 49         pRoot->Inc += c;
 50         return ;
 51     }
 52     pRoot->nSum += c * ( b - a + 1) ;
 53     if( b <= (pRoot->L + pRoot->R)/2)
 54         Add(pRoot->pLeft,a,b,c);
 55     else if( a >= (pRoot->L + pRoot->R)/2 +1)
 56         Add(pRoot->pRight,a,b,c);
 57     else
 58     {
 59         Add(pRoot->pLeft,a,
 60             (pRoot->L + pRoot->R)/2 ,c);
 61         Add(pRoot->pRight,
 62             (pRoot->L + pRoot->R)/2 + 1,b,c);
 63     }
 64 }
 65 long long QuerynSum( CNode * pRoot, int a, int b)
 66 {
 67     if( pRoot->L == a && pRoot->R == b)
 68         return pRoot->nSum +
 69                (pRoot->R - pRoot->L + 1) * pRoot->Inc ;
 70     pRoot->nSum += (pRoot->R - pRoot->L + 1) * pRoot->Inc ;
 71     Add( pRoot->pLeft,pRoot->L,Mid(pRoot),pRoot->Inc);
 72     Add( pRoot->pRight,Mid(pRoot) + 1,pRoot->R,pRoot->Inc);
 73     pRoot->Inc = 0;
 74     if( b <= Mid(pRoot))
 75         return QuerynSum(pRoot->pLeft,a,b);
 76     else if( a >= Mid(pRoot) + 1)
 77         return QuerynSum(pRoot->pRight,a,b);
 78     else
 79     {
 80         return QuerynSum(pRoot->pLeft,a,Mid(pRoot)) +
 81                QuerynSum(pRoot->pRight,Mid(pRoot) + 1,b);
 82     }
 83 }
 84 int main()
 85 {
 86     int n,q,a,b,c;
 87     char cmd[10];
 88     scanf("%d%d",&n,&q);
 89     int i,j,k;
 90     nCount = 0;
 91     BuildTree(Tree,1,n);
 92     for( i = 1; i <= n; i ++ )
 93     {
 94         scanf("%d",&a);
 95         Insert(Tree,i,a);
 96     }
 97     for( i = 0; i < q; i ++ )
 98     {
 99         scanf("%s",cmd);
100         if ( cmd[0] == 'C' )
101         {
102             scanf("%d%d%d",&a,&b,&c);
103             Add( Tree,a,b,c);
104         }
105         else
106         {
107             scanf("%d%d",&a,&b);
108             printf("%I64d\n",QuerynSum(Tree,a,b));
109         }
110     }
111     return 0;
112 }

View Code

转载于:https://www.cnblogs.com/zero-zz/p/4811331.html