3*3 的矩阵，值限定为1-9不重复，已知横竖的和，和一个斜着的值，求这个矩阵，有啥快速的方法没？



x     x     x 11
x     x     x 15
x     x     x 19

16  14   15 15

讨论贴：

http://bbs.csdn.net/topics/391816265

先求横竖斜三行的精确匹配方法：

// puzzl.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"

// puzzle.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <vector>
#include <set>
#include <iostream>

using namespace std;

int result[3][3] = {0};

bool check(int a, int b, int c, int sum)
{
 if((a + b  + c )== sum)
  return true;
 else 
  return false;
}

int lastValue(int a, int b,int sum)
{
 return (sum - a - b);
}

//min max
vector<vector<int> > zuhe(int min,int max,int sum)
{

 vector<vector<int> > quanji;
 for(int i = min;i < max; ++i)
 { 
  for(int j = 1;j < max; ++j)
  {
   if(i + j > sum)
    continue;
   else
   {
    for(int k = 1; k <max ;++k)
    {
     if(check(i,j,k,sum))
     {
      vector<int> ziji;
      ziji.push_back(i);
      ziji.push_back(j);
      ziji.push_back(k);
      if (i!=j&&j!=k&&i!=k)
      {
       quanji.push_back(ziji);
      }
      
      
     }
     else
     {
      continue;
     }
    }
   }
  }

 }

 return quanji;
}

vector<vector<int> > filter_zuhe(int key,vector<vector<int> > temp)
{
 vector<vector<int> >::iterator iter ;
 for(iter = temp.begin(); iter!=temp.end(); )
 {
  if( (*iter)[0] != key)
   iter = temp.erase(iter);
  else
   iter ++ ;
 }

 return temp;
}


bool isOk()
{
	set<int> mySet;

	for(int i = 0;i <3; ++i)
	{
		for(int j = 0;j<3;++j)
		{
			if(result[i][j]>0&&result[i][j]<10)
			{
				mySet.insert(result[i][j]);
			}
		}
	}

	if(mySet.size()!=9)
	{return false;}
	else
	{
		if(mySet.size()==9)
		{
			for(int i = 0;i <3; ++i)
			{
				cout <<endl;
				for(int j = 0;j<3;++j)
				{
					cout<<result[i][j]<<" ";
				}
			}

			return true;
		}
	}
}

void dayin()
{
	for(int i = 0;i <3; ++i)
			{
				cout <<endl;
				for(int j = 0;j<3;++j)
				{
					cout<<result[i][j]<<" ";
				}
			}

	cout<<endl;
}

void qingling()
{
	for(int i = 0;i <3; ++i)
	{
		for(int j = 0;j<3;++j)
		{
			(result[i][j]=0);
		}
	}
}

int _tmain(int argc, _TCHAR* argv[])
{
 vector<vector<int> > quanjiheng = zuhe(1,9,11);
 vector<vector<int> > quanjishu = zuhe(1,9,16);
 vector<vector<int> > quanjixie = zuhe(1,9,15);

 int sizeheng = quanjiheng.size();
 int sizeshu = quanjishu.size();
 int sizexie = quanjixie.size();

 int key = 0,last_key = 0;

 vector<vector<int> > quanjishufilter;
 vector<vector<int> > quanjixiefilter;

 for(int i = 0;i < sizeheng ; ++i)
 {
	   key = quanjiheng[i][0];
	   if (key!=last_key)
	   {
		quanjishufilter = filter_zuhe(key,quanjishu);
		quanjixiefilter = filter_zuhe(key,quanjixie);
	   }
	   last_key = key;
	   //给横行赋值
	  for (int j = 0;j< 3;++j)
	  {
	   result[0][j] = quanjiheng[i][j];
	  
	  }
  
	  int sizeshu = quanjishufilter.size();
	  for (int k = 0; k< sizeshu;++k)
	  {
		  //给竖行赋值
		  for(int j = 0;j<3;++j)
		  {
			   result[j][0] = quanjishufilter[k][j];
		  }

		  int sizexie = quanjixiefilter.size();
		  for(int x = 0; x < sizexie; ++x)
		  {
			  for(int j = 0; j < 3 ;++j)
			  {
				  result[j][j] = quanjixiefilter[x][j];
			  }
				result[2][1] = 14 - result[0][1]-result[1][1];
				result[1][2] = 15-  result[1][0]-result[1][1];
				if(isOk())
				{
					//dayin();
					getchar();
					//return 0;
				}
				else
				{
					
					
				}
		  }

	
	 }

   qingling();
  
 }
 

 getchar();
 system("pause");
 return 0;
}

排列组合的方法：

template <typename T>
void swap(T* array, unsigned int i, unsigned int j)
{
	T t = array[i];
	array[i] = array[j];
	array[j] = t;
}

void FullArray(int* array, size_t array_size, unsigned int index)
{
	if (index >= array_size)
	{
		if ((array[0] + array[1] + array[2] == 11) &&
			(array[3] + array[4] + array[5] == 15) &&
			(array[6] + array[7] + array[8] == 19) && 
			(array[0] + array[3] + array[6] == 16) &&
			(array[1] + array[4] + array[7] == 14) &&
			(array[2] + array[5] + array[8] == 15) &&
			(array[0] + array[4] + array[8] == 15))
		{
			printf("%d,%d,%d\n", array[0], array[1], array[2]);
			printf("%d,%d,%d\n", array[3], array[4], array[5]);
			printf("%d,%d,%d\n\n", array[6], array[7], array[8]);
		}

		return;
	}

	for (unsigned int i = index; i < array_size; ++i)
	{
		swap(array, i, index);

		FullArray(array, array_size, index + 1);

		swap(array, i, index);
	}
}

int main(int argc, char* argv[])
{
	int value[9] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
	
	FullArray(value, 9, 0);
}

模仿人做的方法，任意填写3个空位，保证不同行不同列，或者不在一个斜线。就可以算出来剩下的。