UFOs（树状数组）

Description

Vasya is a ufologist and his duties include observing Unidentified Flying Objects (UFOs) in the part of space bounded by a cube N × N ×N. The cube is divided into cubic sectors 1 × 1 × 1. During the observation, the following events may happen:

• several new UFOs emerge in a certain sector;
• several UFOs disappear in a certain sector;
• Vasya's boss may ask him how many UFOs there are in a part of space consisting of several sectors.

At the moment when Vasya starts his observations there are no UFOs in the whole space.

Input

The first line contains an integer N (1 ≤ N ≤ 128). The coordinates of sectors are integers from 0 to N–1.

Then there are entries describing events, one entry per line. Each entry starts with a number M.

• If M is 1, then this number is followed by four integers x (0 ≤ x < N), y (0 ≤ y < N), z (0 ≤ z < N), K (–20000 ≤ K ≤ 20000), which are coordinates of a sector and the change in the number of UFOs in this sector. The number of UFOs in a sector cannot become negative.
• If M is 2, then this number is followed by six integers x₁, y₁, z₁, x₂, y₂, z₂ (0 ≤ x₁ ≤ x₂ < N, 0 ≤ y₁ ≤ y₂ < N, 0 ≤ z₁ ≤ z₂ < N), which mean that Vasya must compute the total number of UFOs in sectors (x, y, z) belonging to the volume: x₁ ≤ x ≤ x₂, y₁ ≤ y ≤ y₂, z₁ ≤ z≤ z₂.
• If M is 3, it means that Vasya is tired and goes to sleep. This entry is always the last one.

The number of entries does not exceed 100002.

Output

For each query, output in a separate line the required number of UFOs.

Sample Input

input	output
2 2 1 1 1 1 1 1 1 0 0 0 1 1 0 1 0 3 2 0 0 0 0 0 0 2 0 0 0 0 1 0 1 0 1 0 -2 2 0 0 0 1 1 1 3	0 1 4 2

解题思路

这题是三维空间上的，看到userluoxuan用三维树状数组做的，但是三维最后求和的式子太麻烦了，还是二维间接求简单些^_^#

就是将不同的z不同的层单独看，更新时更新z相等这一层，最后求和时求出范围内的每一层再相加。

AC代码

#include<stdio.h>
#include<string.h>
#define maxn 130
int xt[maxn][maxn][maxn] ;
int add, z, n ;
int lowbit( int x )
{
    return x&(-x) ;
}

void update(int x, int y )
{
    int i, j ;
    for(i=x; i<=maxn; i+=lowbit(i))
        for(j=y; j<=maxn; j+=lowbit(j))
        {
            xt[z][i][j] += add ;
            if( xt[z][i][j] < 0 )  xt[z][i][j] = 0 ;            //注意更新，求和时带上第三维
        }
}
int get_sum( int x , int y )
{
    int i, ans=0 ;
    for( i=x; i>0; i-=lowbit(i) )
        for(int j=y; j>0; j-=lowbit(j) )
        {
            ans += xt[z][i][j] ;
        }
    return ans ;
}
int main()
{
    int x, y, a;
    int x1, y1, z1, x2, y2, z2;
    int ans;
    scanf("%d", &n);

    memset(xt, 0, sizeof(xt));
    while( scanf("%d", &a ) && a!=3 )
    {
        ans=0;
        if( a==1 )
        {
            scanf("%d%d%d%d", &x, &y, &z, &add );
            update(x+1, y+1);
        }
        else if( a==2)
        {
            scanf("%d%d%d", &x1, &y1, &z1 );
            scanf("%d%d%d", &x2, &y2, &z2 );
            for(z=z1; z<=z2; z++)
                ans += get_sum(x2+1,y2+1) - get_sum(x1,y2+1) - get_sum(x2+1,y1) + get_sum(x1,y1);

            printf("%d\n",ans);
        }
    }
    return 0;

}