ZOJ 3429 Cube Simulation

链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3429


Cube Simulation
Time Limit: 2 Seconds      Memory Limit: 65536 KB

Here's a cube whose size of its 3 dimensions are all infinite. Meanwhile, there're 6 programs operating this cube:

  • FILL(X,Y,Z): Fill some part of the cube with different values. 

    	memset(cube, 0, sizeof(cube));
	puts("START");
	cnt = 0;
	for (int i = 0; i < X; i++) {
		for (int j = 0; j < Y; j++) {
			for (int k = 0; k < Z; k++) {
				cube[i][j][k] = ++cnt;
			}
		}
	}

  • SWAP1(x1,x2): Swap two sub-cube along the first dimensions. 

    	for (int j = 0; j < Y; j++) {
		for (int k = 0; k < Z; k++) {
			exchange(cube[x1][j][k], cube[x2][j][k]);
		}
	}

  • SWAP2(y1,y2): Swap two sub-cube along the second dimensions. 

    	for (int i = 0; i < X; i++) {
		for (int k = 0; k < Z; k++) {
			exchange(cube[i][y1][k], cube[i][y2][k]);
		}
	}

  • SWAP3(z1,z2): Swap two sub-cube along the third dimensions. 

    	for (int i = 0; i < X; i++) {
		for (int j = 0; j < Y; j++) {
			exchange(cube[i][j][z1], cube[i][j][z2]);
		}
	}

  • FIND(value): Output the value's location, if it exist. 

    	for (int i = 0; i < X; i++) {
		for (int j = 0; j < Y; j++) {
			for (int k = 0; k < Z; k++) {
				if (cube[i][j][k] == value) {
					printf("%d %d %d\n", i, j, k);
				}
			}
		}
	}

  • QUERY(x,y,z): Output the value at (x,y,z). 

    	printf("%d\n", cube[x][y][z]);

We'll give a list of operations mentioned above. Your job is to simulate the program and tell us what does the machine output in progress.

Input

There'll be 6 kind of operations in the input.

  • FILL X Y Z (1 <= X, Y, Z <= 1000) for FILL(X,Y,Z)
  • SWAP1 X1 X2 (0 <= X1, X2 < X) for SWAP1(X1,X2)
  • SWAP2 Y1 Y2 (0 <= Y1, Y2 < Y) for SWAP2(Y1,Y2)
  • SWAP3 Z1 Z2 (0 <= Z1, Z2 < Z) for SWAP3(Z1,Z2)
  • FIND value (value > 0) for FIND(value)
  • QUERY x y z (0 <= x < X, 0 <= y < Y, 0 <= z < Z) for QUERY(x,y,z)

The input will always start with FILL operation and terminate by EOF.

The number of the operations will less than 200,000, while the FILL operation will less than 100.

Output

Simulate all of the operations in order, and print the output of the programs.

Sample Input

FILL 2 3 1
SWAP1 0 1
SWAP2 0 2
SWAP3 0 0
FIND 1
FIND 2
FIND 3
FIND 4
FIND 5
FIND 6
FIND 7
QUERY 0 0 0
QUERY 0 1 0
QUERY 0 2 0
QUERY 1 0 0
QUERY 1 1 0
QUERY 1 2 0

Sample Output

START
1 2 0
1 1 0
1 0 0
0 2 0
0 1 0
0 0 0
6
5
4
3
2
1

Hint

exchange(x,y) means exchange the value of variable x and y.

Because of HUGE input and output, scanf and printf is recommended.

Author: OUYANG, Jialin
Contest: ZOJ Monthly, November 2010


大意——给你一个三维数组。你能够对它进行六种操作。如今给你一系列操作，你的任务就是去模拟它，然后给出在执行过程中，会输出什么东西。


思路——由于数组不能开得太大，所以不能直接去模拟。注意到填充数组时的规律。cube[i][j][k]=k+j*Z+i*Y*Z+1,当中X,Y,Z为填充数组时给出的范围大小。再略微注意一下，能够发现，交换两个值时，仅仅须要交换其对应的索引的值就可以。

那么最后给你索引，你就仅仅须要找到索引的值，然后依照上述公式计算就可以。假设给你值，要你找索引，那你就仅仅须要逆用上述公式找到索引的值，然后再找索引就可以解决。


复杂度分析——时间复杂度:O(X+Y+Z),空间复杂度:O(X+Y+Z)


附上AC代码：


#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
using namespace std;
typedef unsigned int UI;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
const int maxx = 1005;
int cube_x[maxx];
int cube_y[maxx];
int cube_z[maxx]; // 用来存储数组索引
int X, Y, Z; // 数组大小
char str[10]; // 键入命令

void fill(int x, int y, int z); // 填充数组索引
void swap1(int x, int y); // 交换第一个索引
void swap2(int x, int y); // 交换第二个索引
void swap3(int x, int y); // 交换第三个索引
void find(int value); // 找到此值的索引
void query(int x, int y, int z); // 找到此索引的值

int main()
{
	ios::sync_with_stdio(false);
	int x, y, z, value;
	while (~scanf("%s", str))
	{
		if (strcmp("FILL", str) == 0)
		{
			scanf("%d%d%d", &X, &Y, &Z);
			fill(X, Y, Z);
		}
		else if (strcmp("SWAP1", str) == 0)
		{
			scanf("%d%d", &x, &y);
			swap1(x, y);
		}
		else if (strcmp("SWAP2", str) == 0)
		{
			scanf("%d%d", &x, &y);
			swap2(x, y);
		}
		else if (strcmp("SWAP3", str) == 0)
		{
			scanf("%d%d", &x, &y);
			swap3(x, y);
		}
		else if (strcmp("FIND", str) == 0)
		{
			scanf("%d", &value);
			find(value);
		}
		else
		{
			scanf("%d%d%d", &x, &y, &z);
			query(x, y, z);
		}
	}
	return 0;
}

void fill(int x, int y, int z)
{
	printf("START\n");
	for (int i=0; i<x; ++i)
		cube_x[i] = i;
	for (int j=0; j<y; ++j)
		cube_y[j] = j;
	for (int k=0; k<z; ++k)
		cube_z[k] = k;
}

void swap1(int x, int y)
{
	swap(cube_x[x], cube_x[y]);
}

void swap2(int x, int y)
{
	swap(cube_y[x], cube_y[y]);
}

void swap3(int x, int y)
{
	swap(cube_z[x], cube_z[y]);
}

void find(int value)
{
	if (value > X*Y*Z)
		return ;
	int x, y, z;
	x = (value-1)/(Y*Z); // 计算第一个索引
	y = ((value-1)%(Y*Z))/Z; // 计算第二个索引
	z = ((value-1)%(Y*Z))%Z; // 计算第三个索引
	for (int i=0; i<X; ++i)
		if (cube_x[i] == x)
		{
			x = i; // 由于没交换之前索引与数组元素值相等。
				   // 交换之后不一定相等，所以找回原来的
			break;
		}
	for (int j=0; j<Y; ++j)
		if (cube_y[j] == y)
		{
			y = j;
			break;
		}
	for (int k=0; k<Z; ++k)
		if (cube_z[k] == z)
		{
			z = k;
			break;
		}
	printf("%d %d %d\n", x, y, z);
}

void query(int x, int y, int z)
{
	x = cube_x[x];
	y = cube_y[y];
	z = cube_z[z]; // 找到改变后的索引
	int ans = z+y*Z+x*Y*Z+1; // 由于索引从0開始，所以加上1
	printf("%d\n", ans);
}


posted on 2017-08-14 17:12  ljbguanli  阅读(118)  评论(0)  编辑  收藏  举报

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