int dfs(int graph[MAXNODES][MAXNODES],int visited[],int start) {
int stack[MAXNODES];
int top=-1,i;
visited[start]=1;
stack[++top]=start;
while(top!=-1)
{
start=stack[top];
for(i=0;i<MAXNODES;i++) {
if(graph[start][i]&&visited[i]==0) {
stack[++top]=i;
printf("%d-",i);
visited[i]=1;
break;
}
}
if(i==MAXNODES)
top--;
}
return 0;
}
上面的代码在存储为Adjacency Matrix的Graph上实现dfs,我请求一个建议,我应该做些什么改变来知道生成的图是否连接。
答案 0 :(得分:2)
请参阅我之前关于强关联组件的answer问题。
您的dfs写入效率也非常低,因为您重复开始i = 0扫描;你的堆栈应该记住你离开的地方并从那里继续。递归更自然,但如果你有一个有限的调用堆栈大小,那么显式堆栈是最好的(仅适用于大树)。
这是一个递归的dfs。如果你对存储dfs树不感兴趣,你只需在前驱[]中存储1,而不是从你到达它的节点中存储1:
const unsigned MAXNODES=100;
/* predecessor must be 0-initialized by the caller; nodes graph[n] that are
reached from start will have predecessor[n]=p+1, where graph[pred] is the
predecessor via which n was reached from graph[start].
predecessor[start]=MAXNODES+1 (this is the root of the tree; there is NO
predecessor, but instead of 0, I put a positive value to show that it's
reached).
graph[a][b] is true iff there is a directed arc from a->b
*/
void dfs(bool graph[MAXNODES][MAXNODES],unsigned predecessor[]
,unsigned start,unsigned pred=MAXNODES)
{
if (predecessor[start]) return;
predecessor[start]=pred+1;
for (unsigned i=0;i<MAXNODES;++i)
if (graph[start][i])
dfs(graph,predecessor,i,start);
}
这是上面描述的非递归dfs,但为pred
和i
使用相同的堆栈变量(通常,每个局部变量和参数都有一个堆栈变量改变你的递归):
void dfs_iter(bool graph[MAXNODES][MAXNODES],unsigned predecessor[]
,unsigned start)
{
unsigned stack[MAXNODES]; // node indices
unsigned n,pred;
int top=0;
stack[top]=start;
for(;;) {
recurse:
// invariant: stack[top] is the next (maybe reached) node
n=stack[top];
if (!predecessor[n]) { // not started yet
pred=top>0?stack[top-1]:MAXNODES;
//show(n,pred);
predecessor[n]=pred+1;
// the first thing we can reach from n
for (unsigned i=0;i<MAXNODES;++i)
if (graph[n][i] && !predecessor[i]) {
stack[++top]=i; goto recurse; // push
}
}
if (top>0) {
pred=stack[top-1];
// the next thing we can reach from pred after n
for (unsigned i=n+1;i<MAXNODES;++i)
if (graph[pred][i]) {
stack[top]=i; goto recurse; // replace top
}
--top;
} else
return;
}
}
这可以在没有goto的情况下构建(它只是一个命名继续到最外面的循环),但在我看来没有更真实的清晰度。
无论如何,递归调用要简单得多。有Tarjan's strongly connected components algorithm的递归伪代码,你可以直接转录。如果你需要帮助使其非递归(显式堆栈),请询问。
答案 1 :(得分:1)
您需要存储从一个节点到下一个节点行进所生成的边缘。然后,您可以验证图中的所有节点是否通过边连接。
答案 2 :(得分:0)
运行Dijkstra的算法。如果最后您的队列为空并且某些顶点未着色,则表示图形未连接。这是保证线性时间,dfs方法具有二次方的最坏情况分析。