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