Tree Reconstruction
Problem Description
You have just finished a compiler design homework question where you had to find the parse tree of an expression. Unfortunately you left your assignment in the library, but luckily your friend picked it up for you. Instead of e-mailing you the parse tree so that you can rewrite the solution, your friend decides to play a practical joke and sends you just the DFS and BFS trace. Rather than try to redo the entire question you decide to reconstruct the tree.
思路分析
首先这不是一道普通的根据BFS和DFS建树,因此我们需要分析这种树的遍历的性质。普通性质无需多说,这里只说必要的。
对于BFS:
同层的节点BFS编号顺序在DFS中也是按顺序出现的。根据这个我们可以一次遍历算出节点 的深度。
对于DFS:
一个节点的子孙肯定在他后面,在他同层相邻节点的前面。
对于一个节点的子孙,如果他们有的深度恰好等于这个节点的深度+1,那么这些子孙肯定是这个节点的儿子。
有了这些性质我们就可以递归建树了。
代码展现
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<ctime>
#include<iostream>
#include<string>
#include<vector>
#include<list>
#include<deque>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<algorithm>
//#pragma GCC optimize(2)
using namespace std;
#define ll long long
const int maxn=1010;
const int INF=0x7fffffff;
inline void read(int&x){
int data=0,w=1;
char ch=getchar();
while(ch!=‘-‘&&!isdigit(ch))
ch=getchar();
if(ch==‘-‘)
w=-1,ch=getchar();
while(isdigit(ch))
data=10*data+ch-‘0‘,ch=getchar();
x=data*w;
}
void write(int x){
if(x<0)
putchar(‘-‘),x=-x;
if(x>9)
write(x/10);
putchar(‘0‘+x%10);
}
struct vertex{
int depth;
int bfsrank;
int dfsrank;
list<int> edge;
void clear(){
depth=bfsrank=dfsrank=0;
edge.clear();
}
void addedge(int v){
edge.push_back(v);
}
}v[maxn];
int bfsans[maxn],dfsans[maxn];
#define root bfsans[1]
void build(int o,int l,int r){
// clog<<"in o:"<<o<<" l:"<<l<<" r:"<<r<<endl;
for(int i=l;i<=r;++i)
if(v[dfsans[i]].depth==v[o].depth+1)
v[o].addedge(dfsans[i]);
if(v[o].edge.size()){
list<int>::iterator i=v[o].edge.begin(),ed=v[o].edge.end();
--ed;
while(i!=ed){
build(*(i++),v[*(--i)].dfsrank+1,v[*(++i)].dfsrank-1);
}
build(*ed,v[(*ed)].dfsrank+1,r);
}
}
int main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
int n,whodep;
while(cin>>n){
for(int i=1;i<=n;++i)
v[i].clear();
for(int i=1;i<=n;++i){
read(bfsans[i]);
v[bfsans[i]].bfsrank=i;
}
for(int i=1;i<=n;++i){
read(dfsans[i]);
v[dfsans[i]].dfsrank=i;
}
/* clog<<"dfsrank:"<<endl;
for(int i=1;i<=n;i++)
clog<<i<<": "<<v[i].dfsrank<<endl;*/
v[root].depth=0;
whodep=1;
for(int i=2;i<=n;++i){
if(v[bfsans[i]].dfsrank<v[bfsans[i-1]].dfsrank)
++whodep;
v[bfsans[i]].depth=whodep;
}
/* clog<<"depths: "<<endl;
for(int i=1;i<=n;++i)
clog<<i<<": "<<v[i].depth<<endl;*/
build(root,2,n);
for(int i=1;i<=n;++i){
write(i);putchar(‘:‘);putchar(‘ ‘);
if(v[i].edge.size()){
list<int>::iterator j=v[i].edge.begin();
while(j!=v[i].edge.end()){
write(*j);putchar(‘ ‘);
++j;
}
}
putchar(‘\n‘);
}
}
// fclose(stdin);
// fclose(stdout);
return 0;
}
认真看了的同学应该会注意到67行我的奇怪写法。为何不写成build(\*i,v[\*i].dfsrank+1,v[\*(++i)].dfsrank-1);
呢?
这背后就是我写这篇博客的主要原因。上述代码是错的。
万恶之源是函数参数压栈顺序(CALLBACK),他是从右到左压栈的。这样的话上述代码问题就很严重了。我之前就是那样写的,想不到这种注意事项让我碰上了。卡了我半中午啊o(╥﹏╥)o。谁让 std::list
教训惨重......
原文地址:https://www.cnblogs.com/autoint/p/9520696.html