1143. Lowest Common Ancestor

1143. Lowest Common Ancestor (30)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 (<= 1000), the number of pairs of nodes to be tested; and N (<= 10000), 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:

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

Sample Output:

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.

题意就是按照要求找到BST树中U和V点的最近公共祖先并按照要求输出出来

我们可以直接遍历整个表由于BST的先序序列就是元素的插入顺序也就是元素的深度优先搜索的顺序我们可以直接扫描整个表就相当于深度搜索整棵树如果当前点符合输出条件就停下输出

但是后来一直错不知道为什么最后参考了网上的代码才发现原来即使找到了两个点的LCA 也要按照输入顺序先输出U 再输出V

哎。。。这一个错找了好久好久 PAT果然对输入输出一点也不友好。。。

#include<bits/stdc++.h>
using namespace std;
int a[10010];
map<int,bool>Map;
int main()
{int n,m;scanf("%d%d",&n,&m);for(int i=1;i<=m;i++){ scanf("%d",&a[i]);Map[a[i]]=1;}while(n--){int s,e;scanf("%d%d",&s,&e);bool fs=0,fe=0;fs = Map[s],fe = Map[e];if(fs==0&&fe==0)printf("ERROR: %d and %d are not found.\n",s,e);		       else if(fs!=0&&fe==0)printf("ERROR: %d is not found.\n",e); else if(fs==0&&fe!=0)printf("ERROR: %d is not found.\n",s);        else{int ss = s,ee=e;s = min(ss,ee),e = max(ee,ss); for(int now = 1;now<=m;now++){if((s<a[now]&&e>a[now])){printf("LCA of %d and %d is %d.\n",ss,ee,a[now]);//就是这里输出如果是s,e 修改后的元素的输出 就会导致两个测试点出错break;}if(s==a[now]){printf("%d is an ancestor of %d.\n",s,e);break;}if(e==a[now]){printf("%d is an ancestor of %d.\n",e,s);break;}				} }}return 0;
}

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