HDU 3584 HDOJ 3584 Cube ACM 3584 IN HDUCube

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题目描述:

Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 495    Accepted Submission(s): 226

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

1 0 1

 

题目分析 :

      更新区间, 查询一个点, 三维线段树 直接无视.........数据不是很强, 所以 直接暴力就可以过, 过完发现自己的时间 在700MS 左右, 看了下rank,

小A 榜首..... 时间竟然才62MS....果然还是要用三维树状数组来加速啊 . 不过一直没理解 用 树状数组 解决这题的思路, 今早上终于明白了.

画个图先 :

             

好了 , 看代码:

树状数组代码 :

 /*

 std scan_dnum in IsN
ingetcharinEOFininin ingetcharin IsNnum numiningetcharinin
numnuminIsN numnum MAXN matMAXNMAXNMAXN lowMAXN i j k setLow i i MAXN i lowi i i modify x y z i x i MAXN i lowi j y j MAXN j lowj k z k MAXN k lowk
matijk query x y z sum i x i i lowi j y j j lowj k z k k lowk
sum matijk sum
setLow N M scan_d N scan_d M
memset mat mat M x1y1z1x2y2z2 s
scan_d s s
scan_d x1 scan_d y1 scan_d z1
scan_d x2 scan_d y2 scan_d z2
modify x1y1z1
modify x1y1z2
modify x2y1z1
modify x2y1z2
modify x1y2z1
modify x2y2z1
modify x2y2z2
modify x1y2z2
scan_d x1 scan_d y1 scan_d z1
printf query x1y1z1

暴力代码 :

 std node x1 x2 y1 y2 z1 z2cube N M x y z scanfNM EOF cnt M ask
scanfask ask
scanfcubecntx1cubecnty1cubecntz1cubecntx2cubecnty2cubecntz2 cnt
scanfxyz count i i cnt i cubeix1 x x cubeix2
cubeiy1 y y cubeiy2
cubeiz1 z z cubeiz2
count
puts count