正確にk個のエッジを持つソースから宛先までの可能なウォーク

動的計画法を使用して、3Dテーブルを作成する必要があります。行はuの値を指し、列は値vを指し、深さは最初から最後までエッジの数を追跡するために使用されます。

入力と出力

Input:
The adjacency matrix of the graph: The destination vertex is 3. K = 2
0 1 1 1
0 0 0 1
0 0 0 1
0 0 0 0
Output:
There are 2 Possible Walks, from 0 to 3 with 2 edges.

アルゴリズム

numberOdWalks(u, v, k)

入力： 開始頂点u、終了頂点v、エッジの数k。

出力： k個のエッジで可能な歩行の数。

Begin
   define 3D array count of order (n x n x k+1)    //n is number of vertices
   for edge in range 0 to k, do
      for i in range 0 to n-1, do
         for j in range 0 to n-1, do
            count[i, j, edge] := 0
            if edge = 0 and i = j, then
               count[i, j, edge] := 1
            if edge = 1 and (i, j) is connected, then
               count[i, j, edge] := 1
            if edge > 1, then
               for a in range 0 to n, and adjacent with i do
                  count[i, j, edge] := count[i, j, edge] + count[a, j, edge - 1]
               done
         done
      done
   done
   return count[u, v, k]
End

例

#include <iostream>
#define NODE 7
using namespace std;

int graph[NODE][NODE] = {
   {0, 1, 1, 1},
   {0, 0, 0, 1},
   {0, 0, 0, 1},
   {0, 0, 0, 0}
};

int numberOfWalks(int u, int v, int k) {
   int count[NODE][NODE][k+1];

   for (int edge = 0; edge <= k; edge++) {           //for k edges (0..k)
      for (int i = 0; i < NODE; i++) {
         for (int j = 0; j < NODE; j++) {
            count[i][j][edge] = 0;               //initially all values are 0

            if (edge == 0 && i == j)            //when e is 0 and ith and jth node are same, only one path
               count[i][j][edge] = 1;
            if (edge == 1 && graph[i][j])           //when e is 0 and direct path from (i to j), one path
               count[i][j][edge] = 1;
            if (edge >1) {                           //for more than one edges
               for (int a = 0; a < NODE; a++)         // adjacent of source i
                  if (graph[i][a])
                     count[i][j][edge] += count[a][j][edge-1];
            }
         }
      }
   }
   return count[u][v][k];
}

int main() {
   int u = 0, v = 3, k = 2;
   cout << "There are "<< numberOfWalks(u, v, k)<<" Possible Walks, from ";
   cout <<u<<" to "<<v<<" with "<<k<<" edges.";
}

出力

There are 2 Possible Walks, from 0 to 3 with 2 edges.