PAT_甲級_1151 LCA in a Binary Tree (30point(s)) (C++)【LCA倍增法】

目錄

​​1，題目描述​​

​​ 題目大意​​

​​2，思路​​

​​3，AC代碼​​

​​4，解題過程​​

1，題目描述

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.      

 題目大意

找出一棵樹中兩個節點的LCA（lowest common ancestor），注意節點可能不在樹上；

2，思路

​​PAT_甲級_1143 Lowest Common Ancestor (30point(s)) （C++）【BST建構/尋找LCA/倍增法】​​

好像是一模一樣的題目。。。

3，AC代碼

終于體會到手指在鍵盤上飛舞的快感了！

#include<bits/stdc++.h>
using namespace std;
int N, M, in[10005], key;
struct node{
    int key, father, level;
}pre[10005];
void createTree(int root, int left, int right, int father, int level){
    if(left > right) return;
    pre[root] = {pre[root].key, father, level};
    int i = left;
    while(i <= right && pre[root].key != in[i])
        i++;
    createTree(root+1, left, i-1, root, level+1);
    createTree(root+(i-left)+1, i+1, right, root, level+1);
}
int main(){
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif // ONLINE_JUDGE
    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", &key);
        pre[i] = {key, -1, -1};
    }
    createTree(0, 0, N-1, -1, 1);
    while(M--){
        int a, b, A = N, B = N;
        scanf("%d%d", &a, &b);
        for(int i = 0; i < N; i++){
            if(pre[i].key == a)
                A = i;
            if(pre[i].key == b)
                B = i;
        }
        if(A == N && B == N)
            printf("ERROR: %d and %d are not found.\n", a, b);
        else if(A == N)
            printf("ERROR: %d is not found.\n", a);
        else if(B == N)
            printf("ERROR: %d is not found.\n", b);
        else{
            bool flag = true;
            if(pre[A].level < pre[B].level){    
                flag = false;
                swap(A, B);
            }
            while(pre[A].level > pre[B].level)// !!!大于号
                A = pre[A].father;
            if(pre[A].key == pre[B].key)
                printf("%d is an ancestor of %d.\n", flag ? b : a, flag ? a : b);
            else{
                while(pre[A].key != pre[B].key){
                    A = pre[A].father;
                    B = pre[B].father;
                }
                printf("LCA of %d and %d is %d.\n", a, b, pre[A].key);
            }
        }
    }
    return 0;
}      

4，解題過程

一發入魂o(*￣▽￣*)ブ