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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].
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
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 :
/*
Mail to : miyubai@gamil.com
My Blog : www.baiyun.me
Link : http://www.cnblogs.com/MiYu || http://www.cppblog.com/MiYu
Author By : MiYu
Test : 1
Complier : g++ mingw32-3.4.2
Program : HDU_3584
Doc Name : Cube
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
inline bool scan_d(int &num) //整数输入
{
char in;bool IsN=false;
in=getchar();
if(in==EOF) return false;
while(in!='-'&&(in<'0'||in>'9')) in=getchar();
if(in=='-'){ IsN=true;num=0;}
else num=in-'0';
while(in=getchar(),in>='0'&&in<='9'){
num*=10,num+=in-'0';
}
if(IsN) num=-num;
return true;
}
const int MAXN = 105;
int mat[MAXN][MAXN][MAXN];
int low[MAXN];
int i, j, k;
void setLow () {
for ( i = 1; i <= MAXN; ++ i ) low[i] = i & (- i);
}
void modify ( int x, int y, int z ) {
for ( i = x; i <= MAXN; i += low[i] ) {
for ( j = y; j <= MAXN; j += low[j] ) {
for ( k = z; k <= MAXN; k += low[k] )
mat[i][j][k] ^= 1;
}
}
}
int query ( int x, int y, int z ) {
int sum = 0;
for ( i = x; i > 0; i ^= low[i] ) {
for ( j = y; j > 0; j ^= low[j] ) {
for ( k = z; k > 0; k ^= low[k] )
sum += mat[i][j][k];
}
}
return sum & 1;
}
int main ()
{
setLow ();
int N, M;
while ( scan_d ( N ) && scan_d ( M ) ) {
memset ( mat, 0, sizeof ( mat ) );
while ( M -- ) {
int x1,y1,z1,x2,y2,z2;
int s; scan_d ( s );
switch ( s ) {
case 1:
scan_d ( x1 ); scan_d ( y1 ); scan_d ( z1 );
scan_d ( x2 ); scan_d ( y2 ); scan_d ( z2 );
modify ( x1,y1,z1 ); modify ( x1,y1,z2+1 );
modify ( x2+1,y1,z1 ); modify ( x2+1,y1,z2+1 );
modify ( x1,y2+1,z1 ); modify ( x2+1,y2+1,z1 );
modify ( x2+1,y2+1,z2+1 ); modify ( x1,y2+1,z2+1 );
break;
case 0:
scan_d ( x1 ); scan_d ( y1 ); scan_d ( z1 );
printf ( "%d\n", query ( x1,y1,z1 ) );
}
}
}
return 0;
}
暴力代码 :
/*
Mail to : miyubai@gamil.com
My Blog : www.baiyun.me
Link : http://www.cnblogs.com/MiYu || http://www.cppblog.com/MiYu
Author By : MiYu
Test : 1
Complier : g++ mingw32-3.4.2
Program :
Doc Name :
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
struct node {
int x1, x2, y1, y2, z1, z2;
}cube[10010];
int main()
{
int N, M;
int x, y, z;
while ( scanf ( "%d%d",&N,&M ) != EOF ) {
int cnt = 0;
while ( M -- )
{
int ask;
scanf ( "%d", &ask );
switch ( ask ) {
case 1:
scanf ( "%d%d%d%d%d%d", &cube[cnt].x1,&cube[cnt].y1,
&cube[cnt].z1,&cube[cnt].x2,
&cube[cnt].y2,&cube[cnt].z2 );
++ cnt;
break;
case 0:
scanf ( "%d%d%d", &x, &y, &z);
int count = 0;
for ( int i = 0; i != cnt; ++ i )
if ( cube[i].x1 <= x && x <= cube[i].x2 &&
cube[i].y1 <= y && y <= cube[i].y2 &&
cube[i].z1 <= z && z <= cube[i].z2 )
count ^= 1;;
puts ( count ? "1" : "0" );
}
}
}
return 0;
}
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