Nyoj 116 tree Array

Tree array for beginners ..

 

 

This is a line segment tree. We can see from the figure that C [1] = A [1]; C2 = A [1] + A [2]; c [3] = A [3]; C [4] = A [1] + A [2] + A [3] + A [4]; c [5] = A [5];

C [6] = A [5] + A [6]; C [7] = A [7]; c [8] = A [1] + A [2] + A [3] + A [4] + A [5] + A [6] + A [7] + A [8];

We can obtain this rule. If I is an odd number, C [I] = A [I]. Here, 1, 3, 5, and 7 are examples. The even number is related to the power of 2. the maximum number of power of 2 in the I factor, for example, the power of 4 or 2 is 4, therefore, the number of a [4] is a total of 4 numbers, that is, a [1], a [2], a [3], a [4]. When I = 6, the maximum number of factors is 2. Therefore, when I = 8, the three power of 2 is equal to 8, that is, a [1] + ..... the value of a [8. according to the online query, the formula is

C [N] = a (n-a ^ k + 1) + ...... + A [n]; (k indicates the number of digits going from right to 0, that is, the first position where 1 appears, and the start point starts from 0)

Here we can use a function to calculate

Int lowbit (int x)

{

Return X & (-x );

}

How can this function be understood?

When x = 1, the binary representation of X is (assuming 8 digits) 0000 0001

-The binary value of X is 1111 1111.

---> 0000 0001 = 1;

When x = 6, the binary representation of X is 0000 0110.

-The binary value of X is 1111 1010.

---> 0000 0010 = 2;

Here I believe you understand that if you do not understand the binary representation of negative numbers, you can check it by Baidu.

The purpose of this function is to find 2 ^ K (k is the number of numbers that appear from 0 to 0 at the first position where 1 appears)

The following is the sum,

Int sum (int n)

{

Int sum = 0;

While (n> 0)

{

Sum + = C [N];

N = N-lowbit (N );

}

Return sum;

}

How can I understand it? We can think of it as finding the sum of a path, different from the ordinary one-dimensional array increasing or decreasing at a time according to the subscript of 1, here is according to the tree structure, each logn, that is, finding the path (X, x-lowit (x ),.... 1) is the lower mark and

Finally, modify the value of a node.

Void change (int I, int X)

{

While (I <= m) // sum of M nodes

{

C [I] = C [I] + X;

I = I + lowbit (I );

}

}

If the sum above is understood, this step should be quickly understood. If the value of a node is changed, the corresponding points on the corresponding path should also be changed. The condition for the end is to reach the root node.

Let's take a look at this question.

Link: http://acm.nyist.net/JudgeOnline/problem.php? PID = 1, 116

1. If a conventional method is adopted, the time complexity is O (n ^ 2 );

2. The AC code is as follows:

#include<iostream>#include<stdio.h>#include<cstring>using namespace std;int c[1000008],m,Q,t;int lowbit(int x){ return x&(-x);}int Sum(int n){ int sum=0; while(n>0) {  sum=sum+c[n];  n=n-lowbit(n); } return sum;}void change(int i,int x){ while(i<=m) {  c[i]=c[i]+x;  i=i+lowbit(i); }} int main(){ scanf("%d%d",&m,&Q); char key[10]; int start,end; for(int i=1;i<=m;++i) {   scanf("%d",&t);   change(i,t); }    while(Q--)    {  scanf("%s%d%d",key,&start,&end);     if(key[0]=='Q')     printf("%d\n",Sum(end)-Sum(start-1));     else if(key[0]=='A')     change(start,end); } return 0;}

Reference Source: http://www.cnblogs.com/zhangshu/archive/2011/08/16/2141396.html

Http://hi.baidu.com/czyuan_acm/item/b14bff6ab6ffd093c5d249db