1143 Lowest Common Ancestor (30)

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 (≤ 1,000), the number of pairs of nodes to be tested; and N (≤ 10,000), 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:

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

Sample Output:

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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.

题目大意：给出一棵二叉搜索树的前序遍历序列，查询给出值 u 和 v 的共同最低祖先的值。注意输出要求。
分析：先用一个 map 记录出现过的值，这样在读入查询的时候就能检查查询值是否在二叉搜索树中。这道题无需建树，不妨设 u 小于 v，且一定都存在于二叉搜索树中。u 和 v 的最近公共祖先要么是 u 或者 v 本身，要么是一个处于 u 和 v 中间的值。

查找最近公共祖先时，从序列的第一个值val 开始检查（也就是二叉搜索树的根节点），如果 val >= u 且 val <= v，说明就是要找的公共祖先，否则如果 val > u 且 val > v，说明 u 和 v 都在当前节点的右子树，则在前序序列中找到第一个大于 val 的值，继续查找；如果 val < u 且 val < v，说明 u 和 v 都在当前节点的左子树，则在前序序列中找到第一个小于 val 的值，继续查找。这样直到找到 u 或者 v 本身，或者是找到一个 val 处于 u 和 v 中间即为答案。

cpp 复制代码

#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include   <string>
#include   <vector>
#include   <cstdio>
#include    <queue>
#include    <stack>
#include    <ctime>
#include    <cmath>
#include      <map>
#include      <set>
#define INF 0x3fffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
#define NUMBER_OF_THREADS   10
using namespace std;

int find_anc(int nums[],int n,int index,int a,int b)
{
    if(nums[index]>=a&&nums[index]<=b)return nums[index];

    if(nums[index]>a&&nums[index]>b)return find_anc(nums,n,index+1,a,b);
    else if(nums[index]<a&&nums[index]<b)
    {
        int ind=index+1;
        for(;ind<n;++ind)
        {
            if(nums[ind]>=a||nums[ind]>=b)return find_anc(nums,n,ind,a,b);
        }
    }
    return -1;
}

int main(void)
{
    #ifdef test
    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    clock_t start=clock();
    #endif //test

    int m,n,times=1;scanf("%d%d",&m,&n);
    int nums[n+5]={0};
    map<int,int>mp;
    for(int i=0;i<n;++i)
    {
        scanf("%d",&nums[i]);
        if(!mp[nums[i]])mp[nums[i]]=times,times++;
    }

    for(int i=0;i<m;++i)
    {
        int a,b;scanf("%d%d",&a,&b);
        int f1=1,f2=1;
        if(!mp[a])f1=0;
        if(!mp[b])f2=0;
        if(f1==0&&f2)printf("ERROR: %d is not found.\n",a);
        else if(f1&&f2==0)printf("ERROR: %d is not found.\n",b);
        else if(f1==0&&f2==0)printf("ERROR: %d and %d are not found.\n",a,b);
        else
        {

            int ans=0;
            if(a>b)ans=find_anc(nums,n,0,b,a);
            else ans=find_anc(nums,n,0,a,b);
            if(ans!=a&&ans!=b)printf("LCA of %d and %d is %d.\n",a,b,ans);
            else if(ans==a)
            {
                printf("%d is an ancestor of %d.\n",a,b);
            }
            else if(ans==b)
            {
                printf("%d is an ancestor of %d.\n",b,a);
            }
        }
    }


    #ifdef test
    clockid_t end=clock();
    double endtime=(double)(end-start)/CLOCKS_PER_SEC;
    printf("\n\n\n\n\n");
    cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位
    cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位
    #endif //test
    return 0;
}