Cube Summation
Problem Statement :
You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coordinate between y1 and y2 (inclusive) and z coordinate between z1 and z2 (inclusive). Input Format The first line contains an integer T, the number of test-cases. T testcases follow. For each test case, the first line will contain two integers N and M separated by a single space. N defines the N * N * N matrix. M defines the number of operations. The next M lines will contain either 1. UPDATE x y z W 2. QUERY x1 y1 z1 x2 y2 z2 Output Format Print the result for each QUERY. Constrains 1 <= T <= 50 1 <= N <= 100 1 <= M <= 1000 1 <= x1 <= x2 <= N 1 <= y1 <= y2 <= N 1 <= z1 <= z2 <= N 1 <= x,y,z <= N -109 <= W <= 109 Sample Input 2 4 5 UPDATE 2 2 2 4 QUERY 1 1 1 3 3 3 UPDATE 1 1 1 23 QUERY 2 2 2 4 4 4 QUERY 1 1 1 3 3 3 2 4 UPDATE 2 2 2 1 QUERY 1 1 1 1 1 1 QUERY 1 1 1 2 2 2 QUERY 2 2 2 2 2 2 Sample Output 4 4 27 0 1 1
Solution :
Solution in C :
In C ++ :
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<string>
#include<string.h>
#include<cstring>
#include<stack>
#include<queue>
#include<cassert>
#include<cmath>
using namespace std;
#define LL long long int
#define PII pair<int,int>
#define PB push_back
#define MP make_pair
#define INF 1000000000
#define debug(args...) do {cerr << #args << ": "; dbg,args; cerr << endl;} while(0)
LL BIT[110][110][110];
LL old[110][110][110];
void update(int x, int y,int z, LL w){
int i,j,k;
x += 5;
y += 5;
z += 5;
for(i=x;i<110;i+=(i&-i)){
for(j=y;j<110;j+=(j&-j)){
for(k=z;k<110;k+=(k&-k))
BIT[i][j][k] += w;
}
}
}
LL query(int x, int y, int z){
int i,j,k;
LL ret =0;
x += 5;
y += 5;
z += 5;
for(i=x;i>0;i-=(i&-i)){
for(j=y;j>0;j-=(j&-j)){
for(k=z;k>0;k-=(k&-k))
ret += BIT[i][j][k];
}
}
return ret;
}
int main(){
int t,n,m,x,y,z,w;
scanf("%d",&t);
string type;
while(t--){
memset(BIT,0,sizeof(BIT));
memset(old,0,sizeof(old));
cin >> n >> m;
while(m--){
cin >> type;
if(type == "UPDATE"){
scanf("%d %d %d %d",&x,&y,&z,&w);
update(x,y,z,w-old[x][y][z]);
old[x][y][z] = w;
}
else{
int x1,y1,z1,x2,y2,z2;
scanf("%d %d %d %d %d %d",&x1,&y1,&z1,&x2,&y2,&z2);
printf("%Ld\n",
query(x2,y2,z2) -
query(x2,y2,z1-1) -
query(x2,y1-1, z2) -
query(x1-1, y2, z2) +
query(x1-1, y1-1, z2) +
query(x1-1, y2, z1-1) +
query(x2, y1-1, z1-1) -
query(x1-1,y1-1,z1-1)
);
}
}
}
return 0;
}
In Java :
import java.util.Scanner;
public class Solution {
Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
new Solution().ss();
}
private void ss() {
int nrt = sc.nextInt();
for (int i = 0; i < nrt; i++) {
solve();
}
}
long mat[][][];
int n;
private void update(int x, int yy, int zz, long val) {
while (x <= n) {
int y = yy;
while (y <= n) {
int z = zz;
while (z <= n) {
mat[x][y][z] += val;
z += (z & -z);
}
y += (y & -y);
}
x += (x & -x);
}
}
private long sum(int x, int yy, int zz) {
long rez = 0;
while (x > 0) {
int y = yy;
while (y > 0) {
int z = zz;
while (z > 0) {
rez += mat[x][y][z];
z -= (z & -z);
}
y -= (y & -y);
}
x -= (x & -x);
}
return rez;
}
private void solve() {
n = sc.nextInt();
int m = sc.nextInt();
mat = new long[101][101][101];
long[][][] actual = new long[101][101][101];
for (int i = 0; i < m; i++) {
String op = sc.next();
if (op.equals("UPDATE")) {
int x = sc.nextInt(), y = sc.nextInt(), z = sc.nextInt();
long w = sc.nextLong();
// x--; y--; z--;
update(x, y, z, w - actual[x][y][z]);
actual[x][y][z] = w;
} else {
int x1 = sc.nextInt(), y1 = sc.nextInt(), z1 = sc.nextInt();
int x2 = sc.nextInt(), y2 = sc.nextInt(), z2 = sc.nextInt();
// x1--; y1--; z1--;
// x2--; y2--; z2--;
long v1 = sum(x2,y2,z2)- sum(x1-1,y2,z2) - sum(x2,y1-1,z2) + sum(x1-1,y1-1,z2);
long v2 = sum(x2,y2,z1-1) - sum(x1-1,y2,z1-1) - sum(x2,y1-1,z1-1) + sum(x1-1,y1-1,z1-1);
System.out.println(v1 - v2);
}
}
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define max 3001
#define max1 101
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int t,n,m,x1,x2,y1,y2,z1,z2;
int i,j,k,x[max],y[max],z[max];
long W[max];
char str[7];
scanf("%d",&t);
j=0;
while(t--)
{
long mat[max1][max1][max1]={0};
scanf("%d%d",&n,&m);
for(i=0;i<m;i++)
{
scanf("%s",str);
if(str[0]=='U')
{
scanf("%d%d%d%ld",&x[j],&y[j],&z[j],&W[j]);
mat[x[j]][y[j]][z[j]]=W[j];
j++;
}
else
{
scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
long sum=0;
for(k=0;k<j;k++)
{
if(x[k]>=x1 && x[k]<=x2)
{
if(y[k]>=y1 && y[k]<=y2)
{
if(z[k]>=z1 && z[k]<=z2)
{
if(W[k]==mat[x[k]][y[k]][z[k]])
sum+=W[k];
}
}
}
}
printf("%ld\n",sum);
}
}
}
return 0;
}
In Python3 :
numTests = int(input())
#Making the data structure
cube = [];
for i in range(0, 100):
cube.append([]);
for j in range(0, 100):
cube[i].append([])
for k in range(0, 100):
cube[i][j].append(0)
while numTests != 0:
indexList = [[0,0,0,0]]
line = input()
tokens = line.split(' ')
N = int(tokens[0])
numOp = int(tokens[1])
#Updating the data structure according to the queries
while numOp != 0:
query = input()
tokens = query.split(' ')
if(tokens[0] == 'UPDATE'):
x = int(tokens[1]) - 1
y = int(tokens[2]) - 1
z = int(tokens[3]) - 1
w = int(tokens[4])
cube[x][y][z] = w
tempList = []
tempList.append(x)
tempList.append(y)
tempList.append(z)
tempList.append(w)
flag = 0
for index in indexList:
if index[0] == x and index[1] == y and index[2] == z:
index[3] = w
flag = 1
if flag == 0:
indexList.append(tempList)
elif(tokens[0] == 'QUERY'):
x1 = int(tokens[1]) - 1
y1 = int(tokens[2]) - 1
z1 = int(tokens[3]) - 1
x2 = int(tokens[4]) - 1
y2 = int(tokens[5]) - 1
z2 = int(tokens[6]) - 1
sum = 0
for index in indexList:
if index[0] >= x1 and index[0] <= x2:
if index[1] >= y1 and index[1] <= y2:
if index[2] >= z1 and index[2] <= z2:
sum += index[3]
print(sum)
numOp -= 1
numTests -= 1
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