UFOs

1470. UFOs

Time limit: 5.0 second
Memory limit: 64 MB
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 x1y1z1x2y2z2 (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 (xyz) belonging to the volume: x1 ≤ x ≤ x2y1 ≤ y ≤ y2z1 ≤ 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 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

感觉和一维的没什么太大区别,难度也不大,只要理解lowbit就可以轻松写出~~三维,只是比一维多了两个限制条件而已~~
/**三维树状数组**/
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long  ll;
typedef unsigned long long ull;

int dx[4]= {-1,1,0,0};
int dy[4]= {0,0,-1,1}; //up down left right
bool inmap(int x,int y,int n,int m)
{
    if(x<1||x>n||y<1||y>m)return false;
    return true;
}
int hashmap(int x,int y,int m)
{
    return (x-1)*m+y;
}

#define eps 1e-8
#define inf 0x7fffffff
#define debug puts("BUG")
#define lowbit(x) (-x)&x
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#define maxn 200

long long sum[maxn][maxn][maxn]= {0};
int N;

void change(int x,int y,int z,int add)//更新空间中的某一个点
{
    //debug;
    for(int i=x; i<=N; i+=lowbit(i))
        for(int j=y; j<=N; j+=lowbit(j))
            for(int k=z; k<=N; k+=lowbit(k))
            {
                sum[i][j][k]+=add;
                //debug;
            }
    // debug;
}

long long query(int x,int y,int z)//查询从(1,1,1)到(x,y,z)这块立方体区域中ufo的个数
{
    long long ans=0;
    for(int i=x; i>0; i-=lowbit(i))
        for(int j=y; j>0; j-=lowbit(j))
            for(int k=z; k>0; k-=lowbit(k))
                ans+=sum[i][j][k];
    return ans;
}

int main()
{
    int M;
    scanf("%d",&N);
    while(scanf("%d",&M)!=EOF&&M!=3)
    {
        if(M==1)
        {
            int x,y,z,add;
            scanf("%d%d%d%d",&x,&y,&z,&add);
            change(x+1,y+1,z+1,add);
        }
        else if(M==2)
        {
            int x1,x2,y1,y2,z1,z2;
            scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
            long long v= query(x2+1,y2+1,z2+1)
                       - query(x2+1,y2+1,z1) - query(x2+1,y1,z2+1) - query(x1,y2+1,z2+1)
                       + query(x1,y1,z2+1)   + query(x1,y2+1,z1)   + query(x2+1,y1,z1)
                       - query(x1,y1,z1);
/**关键点:要求任意区域的ufo个数,需要将多余的部分减去,多减去的部分重新加回来**/
            printf("%lld\n",v);
        }
    }
    return 0;
}


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