图的深度优先遍历的课本定义是:
受bilibili懒猫老师的讲解启发,我更愿意把图的DFS遍历过程概括为:先遍历图的根节点,再遍历根节点的第一邻接子图,依次遍历完根节点的所有子图。图的根节点就是遍历图时指定的第一个顶点。这样更好记忆其精髓,要点。
递归DFS时调用系统栈。不采用递归的方法,也可以,就自己建立一个栈,存储图里顶点的下标,从栈里依次弹出顶点的顺序,就是DFS遍历图的顺序。
具体过程为,每弹出一个顶点,就修改其对应的visited[]数组的值。对图的遍历,无论DFS还是BFS,都用了visited[]数组,来存储和表明哪些顶点已被访问输出,哪些顶点未被访问输出,防止输出重复顶点。弹出一个顶点后,就把其所有未被输出的邻接点的下标存入栈。依次循环,直至栈为空。
遍历要求不重不漏即可。先遍历根节点的哪个邻接节点也是不确定的。所以DFS遍历结果也是不唯一的。
完整代码如下,先是main函数所在源文件:
#include<iostream>
#include<stdio.h>
using namespace std;
#define NUMVERTEX 6
#define INFINI 65555
struct GraphAdjaMatrix {
char vertexes[NUMVERTEX];
int edges[NUMVERTEX][NUMVERTEX];
int numVertexes;
int numEdges;
};
struct AdjaListNode {
int indexOfVertex;
int weightOfEdge;
AdjaListNode* pt;
};
struct AdjListHead {
char vertex;
AdjaListNode* pt;
};
struct GraphAdjaList {
AdjListHead vertexes[NUMVERTEX];
int numVertexes;
int numEdges;
};
extern void createGraphAdjMatrix(GraphAdjaMatrix &graphAdjMatrix,
int numVertexes,int numEdges,int edges[][NUMVERTEX],char vertexes[]);
extern void createGraphAdjList(GraphAdjaList& graphAdjList,
int numVertexes, int numEdges, int edges[][NUMVERTEX], char vertexes[]);
extern void dispalyGraphAdjMatrix(GraphAdjaMatrix &graphAdjMatrix);
extern void displayGrapgAdjList(GraphAdjaList& graphAdjList);
extern void DFSNoTraverse(GraphAdjaList& graphAdjList,int indexStart);
int main() {
GraphAdjaMatrix graphAdjMatrix ;
GraphAdjaList graphAdjList;
int numEdges = 10;
int edges[][6] = { {0,5,INFINI,7,INFINI,INFINI},
{INFINI,0,4,INFINI,INFINI,INFINI},
{8,INFINI,0,INFINI,INFINI,9},
{INFINI,INFINI,5,0,INFINI,6},
{INFINI,INFINI,INFINI,5,0,INFINI},
{3,INFINI,INFINI,INFINI,1,0} };
char vertexes[] = {'0','1','2','3','4','5'};
createGraphAdjMatrix(graphAdjMatrix,NUMVERTEX,numEdges,edges,vertexes);
createGraphAdjList(graphAdjList,NUMVERTEX,numEdges,edges,vertexes);
dispalyGraphAdjMatrix(graphAdjMatrix);
displayGrapgAdjList(graphAdjList);
cout << endl;
cout << "DFS no traverse : ";
DFSNoTraverse(graphAdjList,0);
return 0;
}
接着是各函数所在源文件:
#include<iostream>
#include<stdio.h>
using namespace std;
#define NUMVERTEX 6
#define INFINI 65555
struct GraphAdjaMatrix {
char vertexes[NUMVERTEX];
int edges[NUMVERTEX][NUMVERTEX];
int numVertexes;
int numEdges;
};
struct AdjaListNode {
int indexOfVertex;
int weightOfEdge;
AdjaListNode* pt;
};
struct AdjListHead {
char vertex;
AdjaListNode* pt;
};
struct GraphAdjaList {
AdjListHead vertexes[NUMVERTEX];
int numVertexes;
int numEdges;
};
void createGraphAdjMatrix(GraphAdjaMatrix &graphAdjMatrix,
int numVertexes, int numEdges, int edges[][NUMVERTEX], char vertexes[]) {
graphAdjMatrix.numVertexes = numVertexes;
graphAdjMatrix.numEdges = numEdges;
for (int i = 0; i < numVertexes; i++)
graphAdjMatrix.vertexes[i] = vertexes[i];
for (int row = 0; row < numVertexes; row++)
for (int column = 0; column < numVertexes; column++)
graphAdjMatrix.edges[row][column] = edges[row][column];
}
void createGraphAdjList(GraphAdjaList &graphAdjList,
int numVertexes, int numEdges, int edges[][NUMVERTEX], char vertexes[]){
graphAdjList.numEdges = numEdges;
graphAdjList.numVertexes = numVertexes;
for (int i = 0; i < numVertexes; i++)
graphAdjList.vertexes[i].pt = NULL;
for (int i = 0; i < numVertexes; i++)
graphAdjList.vertexes[i].vertex = vertexes[i];
AdjaListNode* ptTail = NULL,*ptNew;
int i, j;
for ( i = 0; i < numVertexes; i++)
for (j = 0; j < numVertexes; j++)
if (edges[i][j] != 0 && edges[i][j] != INFINI) {
ptNew = new AdjaListNode;
ptNew->indexOfVertex = j;
ptNew->weightOfEdge = edges[i][j];
if (graphAdjList.vertexes[i].pt == NULL) {
ptNew->pt = NULL;
graphAdjList.vertexes[i].pt = ptNew;
ptTail = ptNew;
}
else {
ptNew->pt = ptTail->pt;
ptTail->pt = ptNew;
ptTail = ptNew;
}
}
}
void dispalyGraphAdjMatrix(GraphAdjaMatrix &graphAdjMatrix) {
cout << "adjacensy matrix :" << endl;
int row,column;
printf("%3c",' ');
for (row = 0; row < graphAdjMatrix.numVertexes; row++)
printf("%3c",graphAdjMatrix.vertexes[row]);
printf("\n");
for (row = 0; row < graphAdjMatrix.numVertexes; row++) {
printf("%-3c", graphAdjMatrix.vertexes[row]);
for (column = 0; column < graphAdjMatrix.numVertexes; column++)
if (graphAdjMatrix.edges[row][column] == INFINI)
printf("%3s", "∞");
else
printf("%3d",graphAdjMatrix.edges[row][column]);
cout << endl;
}
}
void displayGrapgAdjList(GraphAdjaList &graphAdjList) {
cout << "graph adjacency list : " << endl;
AdjaListNode* pt;
int index;
for (int i = 0; i < graphAdjList.numVertexes; i++) {
printf("%2c:",graphAdjList.vertexes[i].vertex);
pt = graphAdjList.vertexes[i].pt;
while (pt != NULL) {
index = pt->indexOfVertex;
printf("%5c(%d)",graphAdjList.vertexes[index].vertex,pt->weightOfEdge);
pt = pt->pt;
}
cout << endl;
}
}
void DFSNoTraverse(GraphAdjaList& graphAdjList, int indexStart) {
bool visited[NUMVERTEX];
for (int i = 0; i < NUMVERTEX; i++)
visited[i] = false;
AdjaListNode* pt;
int stack[NUMVERTEX * 2], top = 0;
stack[top] = indexStart;
while (top >= 0) {
cout << graphAdjList.vertexes[stack[top]].vertex << " ";
visited[stack[top]] = true;
pt = graphAdjList.vertexes[stack[top]].pt;
top--;
while (pt != NULL) {
if (visited[pt->indexOfVertex] == false) {
top++;
stack[top] = pt->indexOfVertex;
}
pt = pt->pt;
}
}
}
测试结果及对应的图如下:
课本给出的DFS非递归的结果是 0 1 2 5 4 3 。我认为都是正确的。
谢谢阅读。