PAT A1151 2019.08.16 【公共祖先】

1151 LCA in a Binary Tree (30 分)

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.

Given any two nodes in a binary tree, 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 (≤ 1,000), the number of pairs of nodes to be tested; and N (≤ 10,000), the number of keys in the binary tree, respectively. In each of the following two lines, N distinct integers are given as the inorder and preorder traversal sequences of the binary tree, respectively. It is guaranteed that the binary tree can be uniquely determined by the input sequences. 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 binary tree, 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
7 2 3 4 6 5 1 8
5 3 7 2 6 4 8 1
2 6
8 1
7 9
12 -3
0 8
99 99

Sample Output:

LCA of 2 and 6 is 3.
8 is an ancestor of 1.
ERROR: 9 is not found.
ERROR: 12 and -3 are not found.
ERROR: 0 is not found.
ERROR: 99 and 99 are not found.

思路分析

根据中序先序遍历序列建树，之后判断公共祖先。与之前有道题类似

刚开始用了 链表树+DFS的方式，但是有一个15分的测试点过不去，只能拿一半分，样例是没问题的

之后参考柳神代码直接用静态树(用了map)

 

柳神代码（30/30）：

//A1151 柳神

#include<iostream>
#include<vector>
#include<map>
using namespace std;
map<int,int> pos;
vector<int> in,pre;

void lca(int inl,int inr,int preRoot,int a,int b)
{
	if(inl>inr)return;
	int inRoot = pos[pre[preRoot]];
	int ain=pos[a],bin=pos[b];
	if(ain<inRoot&&bin<inRoot)
	  lca(inl,inRoot-1,preRoot+1,a,b);
	else if( (ain<inRoot&&bin>inRoot) || (ain>inRoot&&bin<inRoot) )
	  printf("LCA of %d and %d is %d.\n",a,b,in[inRoot]);
	else if(ain>inRoot&&bin>inRoot)
	  lca(inRoot+1,inr,preRoot+1+(inRoot-inl),a,b);
	else if(ain==inRoot)
	  printf("%d is an ancestor of %d.\n",a,b);
	else if(bin==inRoot)
	  printf("%d is an ancestor of %d.\n",b,a);
}

int main()
{
	int m,n,a,b;
	scanf("%d %d",&m,&n);
	in.resize(n+1);
	pre.resize(n+1);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&in[i]);
		pos[in[i]]=i;//值在中序中排第几个 
	}
	for(int i=1;i<=n;i++)scanf("%d",&pre[i]);
	
	for(int i=0;i<m;i++)
	{
		scanf("%d %d",&a,&b);
		if(pos[a]==0 && pos[b]==0)
		  printf("ERROR: %d and %d are not found.\n",a,b);
		else if(pos[a]==0 || pos[b]==0)
		  printf("ERROR: %d is not found.\n",pos[a]==0?a:b);
		else lca(1,n,1,a,b);
	} 
	return 0;
} 

 

我的代码（14/30）：

#include<cstdio>
#include<iostream>
#include<algorithm>
#define MAX 10010
using namespace std;

typedef struct NODE{
	int data;
	struct NODE *left;
	struct NODE *right;
}NODE;

int m,n;
int in[MAX];
int pre[MAX];

NODE *create(int inL,int inR,int preL,int preR)
{
	if(inL>inR)return NULL;
	if(preL>preR)return NULL;
	NODE *node=new NODE;
	node->left=NULL;
	node->right=NULL;
	node->data=pre[preL]; 
	
	int k;
	for(k=inL;k<inR;k++)
	{
		if(in[k]==pre[preL])break; 
	}
	
	int leftNum=k-inL;
	node->left=create(inL,k-1,preL+1,preL+leftNum);
	node->right=create(k+1,preR,preL+leftNum+1,preR);
	return node;
}

int flagA,flagB; // == 1 means can push stack
int stackA[MAX],topA=-1;
int stackB[MAX],topB=-1;
void dfs(NODE *root,int a,int b)
{
	if(root==NULL)return;
	if(flagA==1)stackA[++topA]=root->data;
	if(flagB==1)stackB[++topB]=root->data;
	if(root->data==a)flagA=0;
	if(root->data==b)flagB=0;
	if(flagA==0&&flagB==0)return;

	if(root->left==NULL && root->right == NULL)
	{
		if(flagA==1)topA--;
	    if(flagB==1)topB--;
		return;//leaf
	}
	if(root->left!=NULL)dfs(root->left,a,b);
	if(root->right!=NULL)dfs(root->right,a,b);
	if(flagA==1)topA--;
	if(flagB==1)topB--;
} 

int main()
{
	scanf("%d %d",&m,&n);
	for(int i=0;i<n;i++)scanf("%d",&in[i]);
	for(int i=0;i<n;i++)scanf("%d",&pre[i]);
	NODE *root = new NODE;
	root=create(0,n-1,0,n-1);
	
	for(int i=0;i<m;i++)
	{
		int flaga=1,flagb=1;
		int a,b;
		scanf("%d %d",&a,&b);
		if(a<1||a>n)flaga=0;
		if(b<1||b>n)flagb=0;
		if(flaga==0&&flagb==0)
		  printf("ERROR: %d and %d are not found.\n",a,b);
		else if(flaga==0)
		  printf("ERROR: %d is not found.\n",a);
		else if(flagb==0)
		  printf("ERROR: %d is not found.\n",b);
		else
		{
			flagA=1;flagB=1;
			topA=-1;topB=-1;
			dfs(root,a,b);
			for(int i=0;i<=topA;i++)printf("%d ",stackA[i]);cout<<endl;
			for(int i=0;i<=topB;i++)printf("%d ",stackB[i]);cout<<endl;
            int index=-1;
            int i=0;
            for(i=0;i<=min(topA,topB);i++)
            {
            	if(stackA[i]==stackB[i])
				{
					index++;
					if(i==min(topA,topB))break;
				}
				else break;
			}
//			printf("topA:%d  topB:%d\n",topA,topB);
//			printf("i:%d  index:%d\n",i,index);
			if(index<i)printf("LCA of %d and %d is %d.\n",a,b,stackA[index]);
			else if(topA==topB&&index==topA&&index==topB)
//			      printf("LCA of %d and %d is %d.\n",a,b,stackA[index]);
			      printf("%d is an ancestor of %d.\n",a,b);
//			      printf("%d is an ancestor of %d.\n",b,a);
			else if(index==topA)printf("%d is an ancestor of %d.\n",a,b);
			else if(index==topB)printf("%d is an ancestor of %d.\n",b,a);
			
		}
	}
	
	
}