问题描述：

1143. Lowest Common Ancestor (30)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.

A binary search tree (BST) is recursively defined as a binary tree which has the following properties:

The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
Both the left and right subtrees must also be binary search trees.

Given any two nodes in a BST, you are supposed to find their LCA.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers: M (<= 1000), the number of pairs of nodes to be tested; and N (<= 10000), the number of keys in the BST, respectively. In the second line, N distinct integers are given as the preorder traversal sequence of the BST. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.

Output Specification:

For each given pair of U and V, print in a line "LCA of U and V is A." if the LCA is found and A is the key. But if A is one of U and V, print "X is an ancestor of Y." where X is A and Y is the other node. If U or V is not found in the BST, print in a line "ERROR: U is not found." or "ERROR: V is not found." or "ERROR: U and V are not found.".

Sample Input:

6 8
6 3 1 2 5 4 8 7
2 5
8 7
1 9
12 -3
0 8
99 99

Sample Output:

LCA of 2 and 5 is 3.
8 is an ancestor of 7.
ERROR: 9 is not found.
ERROR: 12 and -3 are not found.
ERROR: 0 is not found.
ERROR: 99 and 99 are not found.

纪念一下3月18日PAT满分×4

认真地写完bst树就能过了，而且求最近公共祖先我是用先对齐结点的深度，再一齐向上搜索的方式求的。。。

AC代码：

#include<bits/stdc++.h>
using namespace std;
struct node
{
	int r;
	int l;
	int d;
	int root;
	int h;
} no;
vector<int> v1,v2;
vector<node> v;
inline void rt(int b1,int b2,int s,int root)
{
	if(s)
	{
		auto itv=find(v1.begin()+b1,v1.begin()+b1+s,v2[b2]);
		int rs=s+b1+v1.begin()-itv-1;
		int ls=itv-v1.begin()-b1;
		if(root<0)
		{
			no.d=*itv;
			no.h=0;
			no.root=0;
			no.l=-1;
			no.r=-1;
			v.push_back(no);
			rt(b1,b2+1,ls,0);
			rt(b1+ls+1,b2+1+ls,rs,0);
		}
		else
		{
			int ro=v.size();
			no.d=*itv;
			no.h=v[root].h+1;
			no.root=root;
			
			if(*itv<v[root].d)
			v[root].l=ro;	
			else
			v[root].r=ro;
			
			no.l=-1;
			no.r=-1;
			v.push_back(no);
			rt(b1,b2+1,ls,ro);
			rt(b1+ls+1,b2+1+ls,rs,ro);	
		}
	}
}
inline int bstfind(int d)
{
	for(int i=0;i>-1;)
	{
		if(d<v[i].d)
		i=v[i].l;
		else if(d>v[i].d)
		i=v[i].r;
		else
		return i;
	}
	return -1;
}
int main()
{
//	freopen("data4.txt","r",stdin);
	ios::sync_with_stdio(false);
	int n,m,c1,c2;
	cin>>n>>m;
	v2.resize(m);
	for(int i=0;i<m;i++)
	{
		cin>>v2[i];
	}
	v1=v2;
	sort(v1.begin(),v1.end());
	rt(0,0,v1.size(),-1);
	for(;n--;)
	{
		cin>>c1>>c2;
		int i1=bstfind(c1);
		int i2=bstfind(c2);
		if(i1<0&&i2<0)
			cout<<"ERROR: "<<c1<<" and "<<c2<<" are not found.\n";
		else if(i1<0)
			cout<<"ERROR: "<<c1<<" is not found.\n";
		else if(i2<0)
			cout<<"ERROR: "<<c2<<" is not found.\n";
		else
		{
			int ii1=i1,ii2=i2;
			for(;ii1!=ii2;)
			{
				if(v[ii1].h>v[ii2].h)
				ii1=v[ii1].root;
				else if(v[ii1].h<v[ii2].h)
				ii2=v[ii2].root;
				else
				{
					ii1=v[ii1].root;
					ii2=v[ii2].root;
				}
			}
			if(ii1==i1)
			cout<<c1<<" is an ancestor of "<<c2<<"."<<endl;
			else if(ii1==i2)
			cout<<c2<<" is an ancestor of "<<c1<<"."<<endl;
			else
			cout<<"LCA of "<<c1<<" and "<<c2<<" is "<<v[ii1].d<<"."<<endl;
		}			
	}
	return 0;
}

PAT Advanced 1143. Lowest Common Ancestor (30)

1143. Lowest Common Ancestor (30)

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