zoukankan      html  css  js  c++  java
  • 1151 LCA in a Binary Tree (30 分)

    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.
    分析:
    根据前序序列和中序序列从根节点开始不断地递归搜索左右子树地根节点,同时不断判断输入地两个孩子相对于当前根地位置。当当前根节点为其中之一或者两孩子分别位于当前根节点地左右子树地时候递归结束。
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 const int maxn = 1e5 + 10;
     4 int pre[maxn], in[maxn];
     5 int prenum[maxn], innum[maxn];
     6 map<int, int> vis;
     7 int n, m;
     8 int a, b;
     9 
    10 int searchroot(int l, int r, int ll, int rr)
    11 {
    12     int rootpos = innum[pre[l]];
    13     int posa = innum[a], posb = innum[b];
    14     if ((rootpos - posa) * (rootpos - posb) <= 0)
    15         return pre[l];
    16     if (rootpos > posa)
    17         return searchroot(l + 1, l + rootpos - ll, ll, rootpos - 1);
    18     else return searchroot(l + rootpos - ll + 1, r, rootpos + 1, rr);
    19 }
    20 
    21 int main()
    22 {
    23     cin >> m >> n;
    24     for (int i = 1; i <= n; i++)
    25         scanf("%d", in + i), vis[in[i]] = 1, innum[in[i]] = i;
    26     for (int i = 1; i <= n; i++)
    27         scanf("%d", pre + i), prenum[pre[i]] = i;
    28     while (m--)
    29     {
    30         cin >> a >> b;
    31         if (!vis[a] && !vis[b]) printf("ERROR: %d and %d are not found.
    ", a, b);
    32         else if (!vis[a]) printf("ERROR: %d is not found.
    ", a);
    33         else if (!vis[b]) printf("ERROR: %d is not found.
    ", b);
    34         else
    35         {
    36             int res = searchroot(1, n, 1, n);
    37             if (res == a)
    38                 printf("%d is an ancestor of %d.
    ", a, b);
    39             else if (res == b)
    40                 printf("%d is an ancestor of %d.
    ", b, a);
    41             else printf("LCA of %d and %d is %d.
    ", a, b, res);
    42         }
    43     }
    44 }
     
  • 相关阅读:
    Daemon Tools手工完全卸载方案
    不要轻易删除/windows/install下文件
    Dumpbin命令的使用
    v4l2 视频捕获
    2瓶4两酒,1个1.5两的酒杯
    n个平面分空间最多可分成多少份
    &#65279导致页面顶部空白一行解决方法
    Base64编码原理分析
    浏览器中“JavaScript解析器”工作原理
    IList转化为DataSet,解决了System.nullable()的问题
  • 原文地址:https://www.cnblogs.com/liuwenhan/p/11863683.html
Copyright © 2011-2022 走看看