算法导论-22.3-6 用栈实现DFS

自己写的一个：

#include <iostream>
#include <stack>
using namespace std;

const int maxV = 100, white = 0, gray = 1, black = 2;
int v, itime;
struct SNode{        //节点
    int color, d, f, num;
    SNode *p;
};
struct SAdj{        //邻接表
    int num;
    SAdj *next;
};
SAdj *adj[maxV];
SNode *vertex[maxV];
stack<SNode*> vertexStack;    //栈

//初始化
void ini(){
for(int i=1; i<=v; i++){
        adj[i] = new SAdj;
        adj[i]->next = NULL;
    }
for(int i=1; i<=v; i++){
        vertex[i] = new SNode;
        vertex[i]->color = white;
        vertex[i]->p = NULL;
        vertex[i]->num = i;
    }
    itime = 0;
}


void dfs(){
    SNode *top;

for(int i=1; i<=v; i++){    //书上的 DFS(G) 部分 （以实现DFS的“多个源顶点”）
        if(vertex[i]->color == white){
            vertex[i]->color = gray;
            vertex[i]->d = ++itime;
            vertex[i]->p = NULL;
            vertexStack.push(vertex[i]);    //新的树根
        }

while(!vertexStack.empty()){
            top = vertexStack.top();
            SAdj *cur = adj[top->num];        //当前节点的邻接表
            while(cur->next != NULL){
if(vertex[cur->next->num]->color == white){
                    vertex[cur->next->num]->color = gray;
                    vertex[cur->next->num]->d = ++itime;
                    vertex[i]->p = top;

                    vertexStack.push(vertex[cur->next->num]);
                    top = vertexStack.top();
                    cur = adj[cur->next->num];
                }
else
                    cur = cur->next;
            }
            top->f = ++itime;        //该节点已完成
            top->color = black;
            vertexStack.pop();
        }
    }

}

void print(){
for(int i=1; i<=v; i++){
        cout << i << " " << vertex[i]->d << " " << vertex[i]->f << endl;
    }
}

int main(){
    cin >> v;

    ini();
int tmpU, tmpV;
while(true){
        cin >> tmpU >> tmpV;
if(tmpU == 0) break;

        SAdj *newNode = new SAdj;
        newNode->num = tmpV;
        newNode->next = adj[tmpU]->next;
        adj[tmpU]->next = newNode;
    }

    dfs();
    print();

return 0;
}

————————————

测试数据：

6
1 4
1 2
2 5
3 6
3 5
4 2
5 4
6 6
0 0

（书 P331）

【另】，参考答案：

DFS(G) //Same as in text
    for each vertex u in V
        color[u] = white
        pi[u] = nil
    time = 0
for each vertex u in V
if color[u] = white
        DFS-Visit(u)

DFS-Visit(u)
    color[u] = gray
    time++    
    d[u] = time
    push(u)

while stack not empty
        u = top()
        isleaf = true
for each v in Adj[u]
if color[v] = white //found new node
           color[v] = grey
           pi[v] = u
           time++
           d[v] = time
           push(v)
           isleaf = false
break

if isleaf = true //u had no more white neighbors
            color[u] = black
            time++
            f[u] = time
            pop() //remove u from stack