hdu 3584 Cube——三维树状数组
火碳黑发表于 2011-05-04 05:54:00
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].
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.
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
1
0
1
分析:
非树状数组莫属，其他数据结构都太帅了，用不着。
先说说一维上的树状数组，什么问题都得成最简单的开始，理解后才能推广到多维或复杂的。
把对某个方体的修改操作看做是对某个方体的操作数+1，那一个位置操作数的奇偶性决定了它的值。
树状数组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......所以对一个区间操作，这个区间首先按树状数组的分区规则，会被分成若干个不相交的区间，在树状数组里对应的就是若干不相交的小树。
比如对1到3操作，会分成分别对[1,2],[3,3]两段的操作；对2到4的操作，可以分成[2,2],[3,3],[4,4]，但是对这种情况，按这种方式去修改树状数组是麻烦的，可以转化成分别对[1,4]操作，再对[1,1]操作，这样[1,1]被修改了两次，所以相当于没修改过(配合这题的奇偶性)，也可处理成对[1,4]的操作数+1，再对[1,1]的操作数-1。
知道每段的信息保存操作次数，可以通过叠加就知道每个位置的操作数。
比如对1到3操作了一次，c[2] = 1(对应[1,2]的区间)，c[3] = 1(对应[3,3]区间)；再对1操作，则c[1] = 1（对应[1,1]区间）。则1的操作数是c[1] + c[2]。
到此还不理解的，建议最好看看树状数组的详细结构，然后画一画就应该清楚多了，表达上的不便请原谅，这个算是比较水的，不过对树状数组的理解是有点用处的，还有就是在树状数组上对方块的切割，例如对对2到4的操作需要处理成[1,4]操作，再对[1,1]操作 ，将这个操作推广到3维。
一维的是区间管区间，二维是方块管理方块，三维是方体管理方体，本质跟一维一样。
#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)//分块更新或操作
{
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)//向上叠加每个父节点的操作次数
{
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
*/
#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)//分块更新或操作
{
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)//向上叠加每个父节点的操作次数
{
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
*/