请参阅我对先前关于强连接组件的问题的 回答。
您的 dfs 在编写时也非常低效,因为您从 i=0 开始重复扫描;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,但如果您有有限的调用堆栈大小,那么显式堆栈是最好的(仅适用于巨大的树)。
这是一个递归的dfs。如果您对存储 dfs 树不感兴趣,则可以将 1 存储在前任[] 中,而不是您从中到达它的节点):
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);
}
这是一个基于上述模式的非递归 dfs,但对pred
and使用相同的堆栈变量i
(通常,对于可以在递归中更改的每个局部变量和参数,您都有一个堆栈变量):
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(它只是一个命名的 continue 到最外层循环),但在我看来没有任何更真正的清晰性。
无论如何,递归调用要简单得多。Tarjan 的强连接组件算法有递归伪代码,您可以直接转录。如果您需要帮助使其成为非递归(显式堆栈),请询问。