如何根据距离矩阵确定节点?

How to determine nodes based on distances matrix?

本文关键字：节点 何根 距离      更新时间：2023-10-16

假设我有一个给定的：

A -> B10

A -> C5

A -> D7.5

B -> C12

B -> D17

C -> D5

然后，我收到一个未排序的输入，如下所示：

K   L   M   N
K   0   10  12  17
L   10  0   5  7.5
M   12  5   0   5
N   17  7.5 5   0

我必须确定(对于任何类型的输入 - 任何顺序(哪个节点(K，L，M & N(实际上是A，B，C和D。

对于上面的示例输入，这里的情况是A is L，B is K，C is M&D is N。

所以我已经开始了一些事情，但我仍然不确定如何继续。下面为我提供了一个 std：：map 输入的哪一行，是给定的行。但是，即使我知道存在这种组合，我也不确定如何知道未知(城市的顺序(。有人可以帮我对输入进行排序以匹配给定的输入吗？

#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
using namespace std;
bool checkForSimilar(vector<double> Vec1, vector<double> Vec2)
{
std::sort(Vec1.begin(), Vec1.end());
std::sort(Vec2.begin(), Vec2.end());
return std::equal(Vec1.begin(), Vec1.end(), Vec1.begin(), Vec2.end());
}
int main()
{
vector<vector<double>> GivenDistances = {  // A   B   C   D
/*A*/ {0,  10, 5,  7.5},
/*B*/ {10, 0,  12, 17 },
/*C*/ {5,  12, 0,  5  },
/*D*/ {7.5,17, 5,  0  }};
vector<vector<double>> InputDistances = {  // K   L   M   N
/*K*/{ 0,  10, 12, 17 },
/*L*/{ 10, 0,  5,  7.5},
/*M*/{ 12, 5,  0,  5  },
/*N*/{ 17, 7.5,5,  0  }};
std::map<int, int> RowMatches;
for (int i = 0; i < InputDistances.size(); i++)
{
for (int j = 0; j < InputDistances[i].size(); j++)
{
// check if current row is any combination if GivenDistances
if (checkForSimilar(InputDistances[i], GivenDistances[j]))
{
RowMatches[i] = j;
}
}
}
// How to order then them??

int pause; 
cin >> pause;
return 0;
}

一个解决问题的函数：

/** Solve problem posed in https://stackoverflow.com/q/52046650/16582
Search for a permuted column in the input matrix
which matches each given column
@param[out] assign  the first node assignment which creates a match
@param[in]  distance  the distances between nodes, given and input
Mean time to find match ( milliseconds )
<pre>
Cities      Search1      Search2
10            20           0.01
100           ???          4
</pre2>
*/
void Find(
cNodeAssign& assign,
cNodeDistance& distance )
{
raven::set::cRunWatch R("Search");
assign.Clear();
// loop over rows in given distances
for( int given = 0; given < distance.Size(); given++ )
{
// loop over rows in input distances
for( int input = 0; input < distance.Size(); input++ )
{
// check if the input row has already been assigned
if( assign.Find( input ) )
continue;
// check if row and column are permutations of each other
if( distance.IsPermutation( given, input ))
{
// found a match
assign.Add( input );
// no need to search further for this row
break;
}
}
}
}

演示函数和执行时间分析的主代码

int main()
{
cout << "Original Problem:n";
vector<vector<double>> GivenDistances =    // A   B   C   D
{
/*A*/ {0,  10, 5,  7.5},
/*B*/ {10, 0,  12, 17 },
/*C*/ {5,  12, 0,  5  },
/*D*/ {7.5,17, 5,  0  }
};
vector<vector<double>> InputDistances =    // K   L   M   N
{
/*K*/{ 0,  10, 12, 17 },
/*L*/{ 10, 0,  5,  7.5},
/*M*/{ 12, 5,  0,  5  },
/*N*/{ 17, 7.5,5,  0  }
};
cNodeDistance dop( GivenDistances, InputDistances );
cNodeAssign assign( dop.Size() );
dop.Display();
Find( assign, dop );
assign.Display();
Demo( 4 );
Demo( 10 );
Timer( 10 );
Timer( 100 );
}

此处提供了用于构建演示应用程序的代码。

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