Cube
Problem Description
Given an N*N*N cube A, whose elements are either 0 or 1. A[i, j, k]
means the number in the i-th row , j-th column and k-th layer.
Initially we have A[i, j, k] = 0 (1 <= i, j, k <= N).
We
define two operations, 1: “Not” operation that we change the A[i, j,
k]=!A[i, j, k]. that means we change A[i, j, k] from 0->1,or
1->0. (x1<=i<=x2,y1<=j<=y2,z1<=k<=z2).
0: “Query” operation we want to get the value of A[i, j, k].
Input
Multi-cases.
First line contains N and M, M lines follow indicating the operation below.
Each operation contains an X, the type of operation. 1: “Not” operation and 0: “Query” operation.
If X is 1, following x1, y1, z1, x2, y2, z2.
If X is 0, following x, y, z.
Output
For each query output A[x, y, z] in one line. (1<=n<=100 sum of m <=10000)
Sample Input
2 5
1 1 1 1 1 1 1
0 1 1 1
1 1 1 1 2 2 2
0 1 1 1
0 2 2 2
Sample Output
題意：在一個(gè)三維長(zhǎng)方體里把0元素改成1或者把1改成0或者詢問(wèn)某個(gè)位置是0還是1.
分析:
非樹(shù)狀數(shù)組莫屬，其他數(shù)據(jù)結(jié)構(gòu)都太帥了，用不著。
先說(shuō)說(shuō)一維上的樹(shù)狀數(shù)組，什么問(wèn)題都得成最簡(jiǎn)單的開(kāi)始，理解后才能推廣到多維或復(fù)雜的。
把對(duì)某個(gè)方體的修改操作看做是對(duì)某個(gè)方體的操作數(shù)+1，那一個(gè)位置操作數(shù)的奇偶性決定了它的值。
樹(shù)狀數(shù)組1位置是管1的，2位置是管1到2的,3是管3的，4是管1到4的......所以對(duì)一個(gè)區(qū)間操作，這個(gè)區(qū)間首先按樹(shù)狀數(shù)組的分區(qū)規(guī)則，會(huì)被分成若干個(gè)不相交的區(qū)間，在樹(shù)狀數(shù)組里對(duì)應(yīng)的就是若干不相交的小樹(shù)。
比如對(duì)1到3操作，會(huì)分成分別對(duì)[1,2],[3,3]兩段的操作；對(duì)2到4的操作，可以分成[2,2],[3,3],[4,4]，但是對(duì)這種情況，按這種方式去修改樹(shù)狀數(shù)組是麻煩的，可以轉(zhuǎn)化成分別對(duì)[1,4]操作，再對(duì)[1,1]操作，這樣[1,1]被修改了兩次，所以相當(dāng)于沒(méi)修改過(guò)(配合這題的奇偶性)，也可處理成對(duì)[1,4]的操作數(shù)+1，再對(duì)[1,1]的操作數(shù)-1。
知道每段的信息保存操作次數(shù)，可以通過(guò)疊加就知道每個(gè)位置的操作數(shù)。
比如對(duì)1到3操作了一次，c[2] = 1(對(duì)應(yīng)[1,2]的區(qū)間)，c[3] = 1(對(duì)應(yīng)[3,3]區(qū)間)；再對(duì)1操作，則c[1] = 1（對(duì)應(yīng)[1,1]區(qū)間）。則1的操作數(shù)是c[1] + c[2]。
到此還不理解的，建議最好看看
樹(shù)狀數(shù)組的詳細(xì)結(jié)構(gòu)，然后畫一畫就應(yīng)該清楚多了，表達(dá)上的不便請(qǐng)?jiān)彛@個(gè)算是比較水的，不過(guò)對(duì)樹(shù)狀數(shù)組的理解是有點(diǎn)用處的，還有就是在樹(shù)狀數(shù)組上對(duì)方塊的切割，例如對(duì)對(duì)2到4的操作需要處理成[1,4]操作，再對(duì)[1,1]操作 ，將這個(gè)操作推廣到3維。
一維的是區(qū)間管區(qū)間，二維是方塊管理方塊，三維是方體管理方體，本質(zhì)跟一維一樣。
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#define maxn 110
using namespace std;
int c[maxn][maxn][maxn];
int n;
void set()
{
int i, j, k;
for (i = 0; i <= n; i++)
{
for (j = 0; j <= n; j++)
{
for (k = 0; k <= n; k++)
{
c[i][j][k] = 0;
}
}
}
}
inline int lowbit(int x)
{
return x & (-x);
}
void update(int x, int y, int z)//分塊更新或操作
{
int i, j, k;
for (i = x; i > 0; i -= lowbit(i))
{
for (j = y; j > 0; j -= lowbit(j))
{
for (k = z; k > 0; k -= lowbit(k))
{
c[i][j][k]++;
}
}
}
}
int sum(int x, int y, int z)//向上疊加每個(gè)父節(jié)點(diǎn)的操作次數(shù)
{
int i, j, k, var = 0;
for (i = x; i <= n; i += lowbit(i))
{
for (j = y; j <= n; j += lowbit(j))
{
for (k = z; k <= n; k += lowbit(k))
{
var += c[i][j][k];
}
}
}
return var;
}
int main()
{
int m, op, x1, y1, z1, x2, y2, z2, ans;
while (scanf("%d%d", &n, &m) - EOF)
{
set();
while (m--)
{
scanf("%d%d%d%d", &op, &x1, &y1, &z1);
if (op == 1)
{
scanf("%d%d%d", &x2, &y2, &z2);
update(x2, y2, z2);
update(x2, y1 - 1, z2);
update(x1 - 1, y2, z2);
update(x1 - 1, y1 - 1, z2);
//--------
update(x2, y2, z1 - 1);
update(x2, y1 - 1, z1 - 1);
update(x1 - 1, y2, z1 - 1);
update(x1 - 1, y1 - 1, z1 - 1);
}
else
{
ans = sum(x1, y1, z1);
printf("%d\n", ans % 2);
}
}
}
return 0;
}
/*
2 100
1 2 2 2 2 2 2
0 1 1 1
1
2 10
1 2 1 2 2
Q 2 2
1 2 1 2 1
Q 1 1
1 1 1 2 1
1 1 2 1 2
1 1 1 2 2
Q 1 1
1 1 1 2 1
Q 2 1
*/