Logic 计算矩形的平铺数
我试图解决这个问题,但找不到解决方案:Logic 计算矩形的平铺数,logic,game-physics,puzzle,Logic,Game Physics,Puzzle,我试图解决这个问题,但找不到解决方案: class Program { // Important note: // The value of masks given here is hard-coded for m == 5. // In a complete solution, you need to calculate the masks for the // actual value of m given. See explanation in answer
class Program
{
// Important note:
// The value of masks given here is hard-coded for m == 5.
// In a complete solution, you need to calculate the masks for the
// actual value of m given. See explanation in answer for more details.
int[] masks = { 0, 3, 6, 12, 15, 24, 27, 30 };
int CountTilings(int n, int m, int s = 0)
{
if (n == 1) { return 1; }
int result = 0;
foreach (int mask in masks)
{
if ((mask & s) == 0)
{
result += CountTilings(n - 1, m, mask);
}
}
return result;
}
public static void Main()
{
Program p = new Program();
int result = p.CountTilings(6, 5);
Console.WriteLine(result);
}
}
// time used : 27 min
#include <set>
#include <vector>
#include <iostream>
using namespace std;
void placement(int n, set< vector <int> > & p){
for (int i = 0; i < n -1 ; i ++){
for (set<vector<int> > :: iterator j = p.begin(); j != p.end(); j ++){
vector <int> temp = *j;
if (temp[i] == 1 || temp[i+1] == 1) continue;
temp[i] = 1; temp[i+1] = 1;
p.insert(temp);
}
}
}
vector<vector<int> > placement( int n){
if (n > 7) throw "error";
set <vector <int> > p;
vector <int> temp (n,0);
p.insert (temp);
for (int i = 0; i < 3; i ++) placement(n, p);
vector <vector <int> > s;
s.assign (p.begin(), p.end());
return s;
}
bool tryput(vector <vector <int> > &board, int current, vector<int> & comb){
for (int i = 0; i < comb.size(); i ++){
if ((board[current][i] == 1 || board[current+1][i]) && comb[i] == 1) return false;
}
return true;
}
void put(vector <vector <int> > &board, int current, vector<int> & comb){
for (int i = 0; i < comb.size(); i ++){
if (comb[i] == 1){
board[current][i] = 1;
board[current+1][i] = 1;
}
}
return;
}
void undo(vector <vector <int> > &board, int current, vector<int> & comb){
for (int i = 0; i < comb.size(); i ++){
if (comb[i] == 1){
board[current][i] = 0;
board[current+1][i] = 0;
}
}
return;
}
int place (vector <vector <int> > &board, int current, vector < vector <int> > & all_comb){
int m = board.size();
if (current >= m) throw "error";
if (current == m - 1) return 1;
int count = 0;
for (int i = 0; i < all_comb.size(); i ++){
if (tryput(board, current, all_comb[i])){
put(board, current, all_comb[i]);
count += place(board, current+1, all_comb) % 10000007;
undo(board, current, all_comb[i]);
}
}
return count;
}
int place (int m, int n){
if (m == 0) return 0;
if (m == 1) return 1;
vector < vector <int> > all_comb = placement(n);
vector <vector <int> > board(m, vector<int>(n, 0));
return place (board, 0, all_comb);
}
int main(){
cout << place(3, 4) << endl;
return 0;
}
给出了一种由N行M列正方形组成的电路板。这块板的瓷砖是覆盖在上面的瓷砖图案。如果:
only tiles of size 1x1 and/or 2x2 are used;
each tile of size 1x1 covers exactly one whole square;
each tile of size 2x2 covers exactly four whole squares;
each square of the board is covered by exactly one tile.
例如,以下图像显示了大小为4行3列的电路板的一些有趣的平铺:
如果电路板上至少有一个正方形在一块瓷砖中覆盖着1x1大小的瓷砖,而在另一块瓷砖中覆盖着2x2大小的瓷砖,则电路板上两个有趣的瓷砖是不同的。例如,上面图像中显示的所有平铺都是不同的
写一个函数
int count_tilings(int N, int M);
给定两个整数N和M,返回大小为N行和M列的电路板上不同有趣的平铺数的10000007模余数
假设:
N is an integer within the range [1..1,000,000];
M is an integer within the range [1..7].
例如,给定N=4和M=3,函数应该返回11,因为有11个大小为4行3列的板的不同有趣的平铺:
对于(4,3),结果是11,对于(6,5),结果是1213。
我尝试了以下方法,但无效:
static public int count_tilings(int N,int M){
int结果=1;
如果((N==1)| |(M==1))返回1;
结果=结果+(N-1)*(M-1);
int max_tiling=(int)((int)(Math.ceil(N/2))*(Math.ceil(M/2));
系统输出打印LN(最大平铺);
对于(int i=2;i=2*i){
int n=i+(n-i);
int k=i;
//System.out.println(“M-1->”+(M-1)+“i->”+i);
System.out.println(“(M-1)^i)->”+(Math.pow((M-1),i));
System.out.println(“n=“+n+”k=“+k”);
系统输出println(组合(n,k));
如果(N-i*2>0){
结果+=数学功率((M-1),i)*组合(n,k);
}否则{
结果+=数学功率((M-1),i);
}
}
如果(M>=2*i){
int n=i+(M-i);
int k=i;
System.out.println(“(N-1)^i)->”+(Math.pow((N-1),i));
System.out.println(“n=“+n+”k=“+k”);
系统输出println(组合(n,k));
如果(M-i*2>0){
结果+=数学功率((N-1),i)*组合(N,k);
}否则{
结果+=数学功率((N-1),i);
}
}
}
返回结果;
}
静态长组合(int n,int k){
/*二项式系数*/
长系数=1;
对于(inti=n-k+1;i由于这是家庭作业,我不会给出完整的解决方案,但我会给你一些提示
首先,这里有一个递归解决方案:
class Program
{
// Important note:
// The value of masks given here is hard-coded for m == 5.
// In a complete solution, you need to calculate the masks for the
// actual value of m given. See explanation in answer for more details.
int[] masks = { 0, 3, 6, 12, 15, 24, 27, 30 };
int CountTilings(int n, int m, int s = 0)
{
if (n == 1) { return 1; }
int result = 0;
foreach (int mask in masks)
{
if ((mask & s) == 0)
{
result += CountTilings(n - 1, m, mask);
}
}
return result;
}
public static void Main()
{
Program p = new Program();
int result = p.CountTilings(6, 5);
Console.WriteLine(result);
}
}
// time used : 27 min
#include <set>
#include <vector>
#include <iostream>
using namespace std;
void placement(int n, set< vector <int> > & p){
for (int i = 0; i < n -1 ; i ++){
for (set<vector<int> > :: iterator j = p.begin(); j != p.end(); j ++){
vector <int> temp = *j;
if (temp[i] == 1 || temp[i+1] == 1) continue;
temp[i] = 1; temp[i+1] = 1;
p.insert(temp);
}
}
}
vector<vector<int> > placement( int n){
if (n > 7) throw "error";
set <vector <int> > p;
vector <int> temp (n,0);
p.insert (temp);
for (int i = 0; i < 3; i ++) placement(n, p);
vector <vector <int> > s;
s.assign (p.begin(), p.end());
return s;
}
bool tryput(vector <vector <int> > &board, int current, vector<int> & comb){
for (int i = 0; i < comb.size(); i ++){
if ((board[current][i] == 1 || board[current+1][i]) && comb[i] == 1) return false;
}
return true;
}
void put(vector <vector <int> > &board, int current, vector<int> & comb){
for (int i = 0; i < comb.size(); i ++){
if (comb[i] == 1){
board[current][i] = 1;
board[current+1][i] = 1;
}
}
return;
}
void undo(vector <vector <int> > &board, int current, vector<int> & comb){
for (int i = 0; i < comb.size(); i ++){
if (comb[i] == 1){
board[current][i] = 0;
board[current+1][i] = 0;
}
}
return;
}
int place (vector <vector <int> > &board, int current, vector < vector <int> > & all_comb){
int m = board.size();
if (current >= m) throw "error";
if (current == m - 1) return 1;
int count = 0;
for (int i = 0; i < all_comb.size(); i ++){
if (tryput(board, current, all_comb[i])){
put(board, current, all_comb[i]);
count += place(board, current+1, all_comb) % 10000007;
undo(board, current, all_comb[i]);
}
}
return count;
}
int place (int m, int n){
if (m == 0) return 0;
if (m == 1) return 1;
vector < vector <int> > all_comb = placement(n);
vector <vector <int> > board(m, vector<int>(n, 0));
return place (board, 0, all_comb);
}
int main(){
cout << place(3, 4) << endl;
return 0;
}
在线查看它的工作情况:
请注意,我添加了一个额外的参数s
。这将存储第一列的内容。如果第一列为空,s=0。如果第一列包含一些填充的正方形,则设置s中的相应位。最初s=0,但当放置一个2 x 2平铺时,这将填充下一列中的一些正方形,这意味着在递归调用中,s将不为零
masks
变量是硬编码的,但在完整的解决方案中,它需要根据m
的实际值进行计算。masks
中存储的值在查看其二进制表示形式时更有意义:
00000
00011
00110
01100
01111
11000
11011
11110
换句话说,它是用m位设置二进制数中的位对的所有方法。您可以编写一些代码来生成所有这些可能性。或者,由于m只有7个可能值,您也可以硬编码掩码的所有7个可能值
然而,递归解决方案有两个严重的问题
如果N
的值较大,则会使堆栈溢出
计算需要指数时间。即使是N
这两个问题都可以通过重写要迭代的算法来解决。保持m
恒定,并初始化n=1
的结果,使s
的所有可能值都成为1
。这是因为如果只有一列,则必须只使用1x1分幅,并且只有一种方法可以做到这一点
现在,您可以使用n=1
的结果来计算n=2
所有可能的s
值。这可以重复,直到您达到n=n
。此算法在与n
相关的线性时间内完成,并且需要恒定的空间。这里是一个递归解决方案:
class Program
{
// Important note:
// The value of masks given here is hard-coded for m == 5.
// In a complete solution, you need to calculate the masks for the
// actual value of m given. See explanation in answer for more details.
int[] masks = { 0, 3, 6, 12, 15, 24, 27, 30 };
int CountTilings(int n, int m, int s = 0)
{
if (n == 1) { return 1; }
int result = 0;
foreach (int mask in masks)
{
if ((mask & s) == 0)
{
result += CountTilings(n - 1, m, mask);
}
}
return result;
}
public static void Main()
{
Program p = new Program();
int result = p.CountTilings(6, 5);
Console.WriteLine(result);
}
}
// time used : 27 min
#include <set>
#include <vector>
#include <iostream>
using namespace std;
void placement(int n, set< vector <int> > & p){
for (int i = 0; i < n -1 ; i ++){
for (set<vector<int> > :: iterator j = p.begin(); j != p.end(); j ++){
vector <int> temp = *j;
if (temp[i] == 1 || temp[i+1] == 1) continue;
temp[i] = 1; temp[i+1] = 1;
p.insert(temp);
}
}
}
vector<vector<int> > placement( int n){
if (n > 7) throw "error";
set <vector <int> > p;
vector <int> temp (n,0);
p.insert (temp);
for (int i = 0; i < 3; i ++) placement(n, p);
vector <vector <int> > s;
s.assign (p.begin(), p.end());
return s;
}
bool tryput(vector <vector <int> > &board, int current, vector<int> & comb){
for (int i = 0; i < comb.size(); i ++){
if ((board[current][i] == 1 || board[current+1][i]) && comb[i] == 1) return false;
}
return true;
}
void put(vector <vector <int> > &board, int current, vector<int> & comb){
for (int i = 0; i < comb.size(); i ++){
if (comb[i] == 1){
board[current][i] = 1;
board[current+1][i] = 1;
}
}
return;
}
void undo(vector <vector <int> > &board, int current, vector<int> & comb){
for (int i = 0; i < comb.size(); i ++){
if (comb[i] == 1){
board[current][i] = 0;
board[current+1][i] = 0;
}
}
return;
}
int place (vector <vector <int> > &board, int current, vector < vector <int> > & all_comb){
int m = board.size();
if (current >= m) throw "error";
if (current == m - 1) return 1;
int count = 0;
for (int i = 0; i < all_comb.size(); i ++){
if (tryput(board, current, all_comb[i])){
put(board, current, all_comb[i]);
count += place(board, current+1, all_comb) % 10000007;
undo(board, current, all_comb[i]);
}
}
return count;
}
int place (int m, int n){
if (m == 0) return 0;
if (m == 1) return 1;
vector < vector <int> > all_comb = placement(n);
vector <vector <int> > board(m, vector<int>(n, 0));
return place (board, 0, all_comb);
}
int main(){
cout << place(3, 4) << endl;
return 0;
}
//使用时间:27分钟
#包括
#包括
#包括
使用名称空间std;
无效位置(int n,set&p){
对于(int i=0;i7)抛出“错误”;
设p;
向量温度(n,0);
p、 插入(温度);
对于(inti=0;i<3;i++)位置(n,p);
向量s;
s、 赋值(p.begin(),p.end());
返回s;
}
bool tryput(矢量和线路板、整数电流、矢量和梳状){
对于(int i=0;i&所有梳状){
int m=board.size();
如果(当前>=m)抛出“错误”;
如果(当前==m-1)返回1;
整数计数=0;
对于(int i=0;i