递归分治实现比赛日程安排（仅对2…

    本来是很简单的一个东西，根本没有必要写下来，不过，鉴于是第一次完全使用Ubuntu下面的基础工具（vim、makefile、g++）实现和调试的，所以，贴点膏药：

    test.cpp：

#include <iostream>

using namespace std;

#define MaxLength 1000

void Table(int **A, int start, int end);

void Merge(int **A, int **B, int **C, int n);

int main()

{

int **A;

int n;

int i, j;

//initialize Matrix A

A = new int *[MaxLength];

for(i = 0; i < MaxLength; i++)

{

A[i] = new int[MaxLength];

}

cout<<"Please input the number of competitor(should be power of 2):"<<endl;

cin>>n;

Table(A, 0,  n - 1);

for(i = 0; i < n; i++)

{

for(j = 0; j < n; j++)

{

cout<<A[i][j]<<" ";

}

cout<<endl;

}

return 0;

}

void Table(int **A, int start, int end)

{

if(start == end - 1)

{

for(int i = 0; i < 2; i++)

for(int j = 0; j < 2; j++)

if(i == j)

A[i][j] = start + 1;

else

A[i][j] = end + 1;

}

else

{

int half = (end - start) / 2;

int **B = new int *[half + 1];

int **C = new int *[half + 1];

for(int i = 0; i < half + 1; i++)

{

B[i] = new int[half + 1];

C[i] = new int[half + 1];

}

Table(B, start, start + half);

Table(C, end - half, end);

Merge(A, B, C, half + 1);

for(int i = 0; i < half; i++)

{

delete B[i];

delete C[i];

}

delete B;

delete C;

}

}

void Merge(int **A, int **B, int **C, int n)

{

int i, j;

for(i = 0; i < n; i++)

for(j = 0; j < n; j++)

{

//diagonal elments

A[i][j] = B[i][j];

A[i + n][j + n] = B[i][j];

//anti-diagonal elements

A[i][j + n] = C[i][j];

A[i + n][j] = C[i][j];

}

}

    Makefile：

test.out:test.o

g++ -o test.out test.o

test.o:test.cpp

g++ -c test.cpp

    运行结果：

Please input the number of competitor(should be power of 2):

8

1 2 3 4 5 6 7 8 

2 1 4 3 6 5 8 7 

3 4 1 2 7 8 5 6 

4 3 2 1 8 7 6 5 

5 6 7 8 1 2 3 4 

6 5 8 7 2 1 4 3 

7 8 5 6 3 4 1 2 

8 7 6 5 4 3 2 1