PAT A1151 LCA in Binary Tree
利用树的前序和中序递归判定最小公共祖先~
直接根据两个序列递归处理~
#include<bits/stdc++.h>
using namespace std;
const int maxn=;
int N,M;
int pre[maxn],in[maxn];
unordered_map<int,int> pos;
void lca (int inL,int inR,int preRoot,int a,int b) {
if (inL>inR) return;
int inRoot=pos[pre[preRoot]];
int aIn=pos[a];
int bIn=pos[b];
if ((aIn>inRoot&&bIn<inRoot)||(aIn<inRoot&&bIn>inRoot))
printf ("LCA of %d and %d is %d.\n",a,b,in[inRoot]);
else if (aIn>inRoot&&bIn>inRoot)
lca (inRoot+,inR,preRoot+inRoot-inL+,a,b);
else if (aIn<inRoot&&bIn<inRoot)
lca (inL,inRoot-,preRoot+,a,b);
else if (aIn==inRoot)
printf ("%d is an ancestor of %d.\n",a,b);
else if (bIn==inRoot)
printf ("%d is an ancestor of %d.\n",b,a);
}
int main () {
scanf ("%d %d",&M,&N);
for (int i=;i<=N;i++) {
scanf ("%d",&in[i]);
pos[in[i]]=i;
}
for (int i=;i<=N;i++)
scanf ("%d",&pre[i]);
int a,b;
for (int i=;i<M;i++) {
scanf ("%d %d",&a,&b);
if (pos[a]==&&pos[b]==)
printf ("ERROR: %d and %d are not found.\n",a,b);
else if (pos[a]==)
printf ("ERROR: %d is not found.\n",a);
else if (pos[b]==)
printf ("ERROR: %d is not found.\n",b);
else lca (,N,,a,b);
}
return ;
}
也可以根据树建图,跑lca算法
#include<bits/stdc++.h>
using namespace std;
const int maxn=;
struct node {
int data;
node * left;
node * right;
};
int M,N;
int pre[maxn],in[maxn];
unordered_map<int,int> pos;
int father[maxn*];
int depth[maxn*];
node * create (int preL,int preR,int inL,int inR) {
if (preL>preR) return NULL;
node * root=new node;
root->data=pre[preL];
int k;
for (k=inL;k<=inR;k++)
if (in[k]==pre[preL]) break;
int numLeft=k-inL;
root->left=create(preL+,preL+numLeft,inL,k-);
if (root->left!=NULL) father[root->left->data]=root->data;
root->right=create(preL+numLeft+,preR,k+,inR);
if (root->right!=NULL) father[root->right->data]=root->data;
return root;
}
void bfs (node * root) {
queue<node *> q;
q.push(root);
depth[root->data]=;
while (!q.empty()) {
node * now=q.front();
q.pop();
if (now->left) {q.push(now->left);depth[now->left->data]=depth[now->data]+;}
if (now->right) {q.push(now->right);depth[now->right->data]=depth[now->data]+;}
}
}
void lca (int u,int v) {
int tu=u;
int tv=v;
while (depth[tu]<depth[tv]) {
tv=father[tv];
}
while (depth[tu]>depth[tv]) {
tu=father[tu];
}
while (tu!=tv) {
tu=father[tu];
tv=father[tv];
}
if (tu==u) printf ("%d is an ancestor of %d.\n",u,v);
else if (tu==v) printf ("%d is an ancestor of %d.\n",v,u);
else printf ("LCA of %d and %d is %d.\n",u,v,tu);
}
int main () {
scanf ("%d %d",&M,&N);
for (int i=;i<=N;i++) father[i]=i;
for (int i=;i<=N;i++) {
scanf ("%d",&in[i]);
pos[in[i]]=i;
}
for (int i=;i<=N;i++)
scanf ("%d",&pre[i]);
node * root=create(,N,,N);
bfs(root);
int u,v;
for (int i=;i<M;i++) {
scanf ("%d %d",&u,&v);
if (pos[u]==&&pos[v]==) printf ("ERROR: %d and %d are not found.\n",u,v);
else if (pos[u]==||pos[v]==) printf ("ERROR: %d is not found.\n",pos[u]==?u:v);
else lca (u,v);
}
return ;
}
PAT A1151 LCA in Binary Tree的更多相关文章
- PAT A1151 LCA in a Binary Tree (30 分)——二叉树,最小公共祖先(lca)
The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U ...
- PAT A1102 Invert a Binary Tree (25 分)——静态树,层序遍历,先序遍历,后序遍历
The following is from Max Howell @twitter: Google: 90% of our engineers use the software you wrote ( ...
- PAT 甲级 1110 Complete Binary Tree
https://pintia.cn/problem-sets/994805342720868352/problems/994805359372255232 Given a tree, you are ...
- PAT 1102 Invert a Binary Tree[比较简单]
1102 Invert a Binary Tree(25 分) The following is from Max Howell @twitter: Google: 90% of our engine ...
- PAT甲级——1110 Complete Binary Tree (完全二叉树)
此文章同步发布在CSDN上:https://blog.csdn.net/weixin_44385565/article/details/90317830 1110 Complete Binary ...
- PAT 1102 Invert a Binary Tree
The following is from Max Howell @twitter: Google: 90% of our engineers use the software you wrote ( ...
- PAT甲级——A1110 Complete Binary Tree【25】
Given a tree, you are supposed to tell if it is a complete binary tree. Input Specification: Each in ...
- PAT Advanced 1110 Complete Binary Tree (25) [完全⼆叉树]
题目 Given a tree, you are supposed to tell if it is a complete binary tree. Input Specification: Each ...
- PAT_A1151#LCA in a Binary Tree
Source: PAT A1151 LCA in a Binary Tree (30 分) Description: The lowest common ancestor (LCA) of two n ...
随机推荐
- 你了解getBoundingClientRect()?
理解:getBoundingClientRect用于获取某个元素相对于视窗的位置集合.集合中有top, right, bottom, left等属性. 1.语法:这个方法没有参数. rectObjec ...
- 洛谷 P2058 海港(模拟)
题目链接:https://www.luogu.com.cn/problem/P2058 这是一道用手写队列模拟的一道题,没有什么细节,只是注意因为数不会很大,所以直接用数作为数组下标即可,不用用map ...
- Win10上面安装vmware,并在Vmware上面安装Ubuntu
一.安装vmware vmware安装包 链接:https://pan.baidu.com/s/178IOOuMOcotSrr6omIAM_A 提取码:c7ba vmware激活码 链接:https: ...
- 唠唠C++二级指针、二维数组、指针数组、数组指针等的区分
今天看c++primer第六章,有这部分的内容,脑子有点糊涂了,看了几篇博客,自己敲了下,记录一下备忘. 二级指针: int **p; 二维数组: int p[10][10]; char q[10][ ...
- 红帽RHCE培训-课程2笔记内容
1 kickstart自动安装 已安装系统中,在root下述目录会自动生成kickstart配置文件 ll ~/anaconda-ks.cfg 关键配置元素注释,详见未精简版 创建Kickstart配 ...
- debug assertion failed问题解决
运行过程中出现上述问题,后来发现是vector越界问题....解决办法:不要越界就好了...
- postgresql数据库利用函数返回查询结果集
- Servlet映射细节
Servlet的映射细节: 1):一个Servlet程序(Web组件),可以配置多个<url-pattern>,表示一个Servlet有多个资源名称. <servlet-mappin ...
- ztree-拖拽(排序树)
<!DOCTYPE html> <HTML> <HEAD> <TITLE> ZTREE DEMO - beforeDrag / onDrag / bef ...
- HashMap遍历,取出key和value
HashMap的遍历有两种常用的方法,那就是使用keyset及entryset来进行遍历,在用keySet(key)取value时候,需要key 第一种: Map map = new HashMap( ...