UFOs

来源：互联网发布：office2011for mac 编辑：程序博客网时间：2024/06/06 03:47

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

题目大意：

描述：Vasya 不明飞行物研究家，他的职责是观察UFO在一个N × N ×N的空间里的存在情况，这个空间被分成1 × 1 × 1的小区域，在这些小区域中会发生下列事件：

一些小区域会出现一些新的UFO；

一些UFO会在一些小区域消失；

Vasya 的上司会询问他某个区域的UFO有多少；

当Vasya开始观察UFO时，在整个区域内没有UFO。

输入：

第一行包含一个整数N(1 ≤ N ≤ 128).。区域的坐标是从0到N-1。

然后又多行描述，每行的第一个数是M。

如果M是1，接着就有四个整数，分别为 x (0 ≤ x < N), y (0 ≤ y < N), z (0 ≤ z < N), K (–20000 ≤ K ≤ 20000)，x，y，z表示UFO出现的小区域，K表示UFO出现的数量。

如果M是2，接着就有六个整数 x1, y1, z1, x2, y2, z2 (0 ≤ x1 ≤ x2 < N, 0 ≤ y1 ≤ y2 < N, 0 ≤ z1 ≤ z2 < N)，（x1，y1，z1）与（x2，y2，z2）分别表示所求区域的体对角线上的点。

如果M是3，那么Vasya就可以结束任务，回家睡觉了。

输出：

对Vasya上司的每句询问，都输出当前该区域UFO的数量，并且结果占一行。

本题可以利用三维树状数组，其实和一维树状数组，二维树状数组差不多，只是在更新（update）函数和求和（SUM）函数中相对一维、二维树状数组多了两、一个for循环而已，在最后的调用SUM函数是稍复杂些罢了。

代码如下：

//三维树状数组#include<stdio.h>#include<string.h>#define max 130int N;//三维树状数组（s[ ][ ][ ]）用来储存出现UFO的数量，由于数量值较大，采用long long型long long s[max][max][max]={0};int lowbit(int x){    return x&(-x);}//更新函数，对树状数组某一元素（x，y，z）加减某一num值void update(int x,int y,int z,int num){    int i,j,k;    for(i=x;i<=N;i+=lowbit(i))    {        for(j=y;j<=N;j+=lowbit(j))        {            for(k=z;k<=N;k+=lowbit(k))                s[i][j][k]+=num;        }    }}//求和函数，对树状数组某区域内的元素求和，注意数值较大，采用long long型long long SUM(int x,int y,int z){    int i,j,k;    long long ans=0;    for(i=x;i>0;i-=lowbit(i))    {        for(j=y;j>0;j-=lowbit(j))        {            for(k=z;k>0;k-=lowbit(k))                ans+=s[i][j][k];        }    }    return ans;}int main(){    long long sum;    int x,y,z,x1,x2,y1,y2,z1,z2,M,K;    scanf("%d",&N);    while(scanf("%d",&M)&&M!=3)    {        if(M==1)        {            //输入要更新的元素以及更新的值            scanf("%d%d%d%d",&x,&y,&z,&K);            /*对相应元素更新，注意树状数组是            从1开始储存的，输入的元素是从0开始的*/            update(x+1,y+1,z+1,K);        }        if(M==2)        {            //输入需要求和的区域            scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);            //求和            sum=SUM(x2+1,y2+1,z2+1)-SUM(x2+1,y2+1,z1)-SUM(x2+1,y1,z2+1)-SUM(x1,y2+1,z2+1)+SUM(x1,y1,z2+1)+SUM(x1,y2+1,z1)+SUM(x2+1,y1,z1)-SUM(x1,y1,z1);            printf("%lld\n",sum);        }    }    return 0;}

0 0