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从源到目的地的可能步行，恰好有 k 条边

2023-06-21 06:38:03 78

给出了一个有向图。还给出了另外两个顶点u和v，u是起始顶点，v是结束顶点。我们的任务是找到从顶点u到顶点v且恰好有k条边的多次步行。算法中还提供了k的值。

通过使用动态规划，我们需要创建一个3D表，其中行将指向u的值，列将指向值v，深度将用于跟踪从头到尾的边数。

输入和输出

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:
有 2 Possible Walks, from 0 to 3 with 2 edges.

算法

numberOdWalks(u, v, k)

输入： 起始顶点u，结束顶点v，边数k。

输出：具有k条边的可能步行数。

Begin
   define 3D array count of order (n x n x k+1)    //n是顶点数
   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

示例

#include 
#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++) {           //对于k条边(0..k)
      for (int i = 0; i < NODE; i++) {
         for (int j = 0; j < NODE; j++) {
            count[i][j][edge] = 0;               //最初所有值都是0

            if (edge == 0 && i == j)            //当e为0且第i个和第j个节点相同时，只有一条路径
               count[i][j][edge] = 1;
            if (edge == 1 && graph[i][j])           //当e为0并且从(i到j)的直接路径时，一条路径
               count[i][j][edge] = 1;
            if (edge >1) {                           //对于多个边缘
               for (int a = 0; a < NODE; a++)         //与源i相邻
                  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 << "有 "<< numberOfWalks(u, v, k)<<" Possible Walks, from ";
   cout <输出结果有 2 Possible Walks, from 0 to 3 with 2 edges.

返回顶部
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czq8825@qq.com

从源到目的地的可能步行，恰好有 k 条边

输入和输出

算法

示例

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