使用以下代碼，當我調用該bfs方法時，得到的結果是123465247。我應該如何聲明和使用visited變量，以使輸出變為1234657？class Graph{ public int[] graph1VertexList = new int[] {1,2,3,4,5,6,7}; public int[,] graph1=new int[7,7]; public Graph() { //for the graph1 graph1[0, 1] = 1; graph1[0, 2] = 1; graph1[1, 3] = 1; graph1[2, 5] = 1; graph1[3, 4] = 1; graph1[4, 1] = 1; graph1[5, 3] = 1; graph1[5, 6] = 1; } public void bfs() { Console.Write(graph1VertexList[0]);//print the first value for(int i = 0; i < graph1VertexList.Length; i++) { for(int z = 0; z < graph1VertexList.Length; z++) { if (graph1[i, z] == 1) { Console.Write(graph1VertexList[z]); } } } }}
1 回答
jeck貓
TA貢獻1909條經驗 獲得超7個贊
您應該引入一個布爾數組，以指示某個頂點是否已經被訪問。此外，應檢查數組是否遵循邊并在更新后進行更新。這可以如下進行。
public int[] graph1VertexList = new int[] { 1, 2, 3, 4, 5, 6, 7 };
public int[,] graph1 = new int[7, 7];
public bool[] visited = new bool[7]; // new array, initialized to false
public void bfs()
{
Console.Write(graph1VertexList[0]);//print the first value
for (int i = 0; i < graph1VertexList.Length; i++)
{
for (int z = 0; z < graph1VertexList.Length; z++)
{
if (graph1[i, z] == 1 && !visited[z]) // check for visit
{
visited[z] = true; // mark as visited
Console.Write(graph1VertexList[z]);
}
}
}
}
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