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

inputoutput

22 1 1 1 1 1 11 0 0 0 11 0 1 0 32 0 0 0 0 0 02 0 0 0 0 1 01 0 1 0 -22 0 0 0 1 1 13

0142

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 130int 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;}