1151 1151 LCA in a Binary Tree PAT 甲级

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.

题目大意：给出中序序列和先序序列，再给出两个点，求这两个点的最近公共祖先～

分析：不用建树～已知某个树的根结点，若a和b在根结点的左边，则a和b的最近公共祖先在当前子树根结点的左子树寻找，如果a和b在当前子树根结点的两边，在当前子树的根结点就是a和b的最近公共祖先，如果a和b在当前子树根结点的右边，则a和b的最近公共祖先就在当前子树的右子树寻找。中序加先序可以唯一确定一棵树，在不构建树的情况下，在每一层的递归中，可以得到树的根结点，在此时并入lca算法可以确定两个结点的公共祖先～

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注意：题目中有大量数据读入，用cin会超时；

        有节点编号小于零，不能用数组（下标不能小于零，选用map）

 

#include<iostream>
#include<vector>
#include<map>
using namespace std;
int n, m;
map<int,int > pos;
vector<int>in, pr;
void dfs(int l,int r,int prroot,int u,int v)
{
    int inroot = pos[pr[prroot]];
    if ((pos[u]<inroot && pos[v]>inroot) || (pos[u] > inroot && pos[v] < inroot)) {
        printf("LCA of %d and %d is %d.
",u,v,in[inroot]);
    }
    else if (pos[v] < inroot && pos[u] < inroot) {
        dfs(l,r-1,prroot+1, u, v);
    }
    else if (pos[v] > inroot && pos[u] > inroot) {
        dfs(inroot+1,r,prroot +1 +inroot-l, u, v);
    }
    else if (u == in[inroot] || v == in[inroot]) {
        printf("%d is an ancestor of %d.
",u == in[inroot]?u:v,u == in[inroot]?v:u);
    }
}
int main()
{
    cin >> m >> n; //题目中有大量的数据读入，用cin会超时
    scanf("")
    in.resize(n + 1), pr.resize(n+1);
        for (int i = 1; i <= n; i++)
        cin >> in[i], pos[in[i]] = i;
    for (int i = 1; i <= n; i++)
        cin >> pr[i];
    while (m--) {
        int u, v;
        cin >> u >> v;
        if (pos[u] == 0 && pos[v] == 0) {

            printf("ERROR: %d and %d are not found.
",u,v);
            continue;
        }
        else if (0 == pos[u] || 0 == pos[v]) {
            printf("ERROR: %d is not found.
",pos[u] == 0?u:v);
            continue;
        }    
        dfs(1,n,1, u, v);
    }
    getchar();
    getchar();
    return 0;
}