1143. Lowest Common Ancestor (30)

時間限制200 ms
內(nèi)存限制65536 kB
代碼長度限制16000 B
判題程序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.

題意就是按照要求找到BST樹中U和V點的最近公共祖先 并按照要求輸出出來

我們可以直接遍歷整個表 由于BST的先序序列 就是元素的插入順序 也就是元素的深度優(yōu)先搜索的順序 我們可以直接掃描整個表 就相當(dāng)于深度搜索整棵樹 如果當(dāng)前點符合輸出條件就停下輸出

但是后來一直錯不知道為什么 最后參考了網(wǎng)上的代碼才發(fā)現(xiàn) 原來即使找到了兩個點的LCA 也要按照輸入順序 先輸出U 再輸出V

哎。。。這一個錯找了好久好久 PAT果然對輸入輸出一點也不友好。。。

#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]);//就是這里輸出如果是s,e 修改后的元素的輸出 就會導(dǎo)致兩個測試點出錯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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