zoukankan      html  css  js  c++  java
  • Day6

    Let us consider a special type of a binary search tree, called a cartesian tree. Recall that a binary search tree is a rooted ordered binary tree, such that for its every node x the following condition is satisfied: each node in its left subtree has the key less then the key of x, and each node in its right subtree has the key greater then the key of x.
    That is, if we denote left subtree of the node x by L(x), its right subtree by R(x) and its key by kx then for each node x we have
    • if y ∈ L(x) then ky < kx
    • if z ∈ R(x) then kz > kx

    The binary search tree is called cartesian if its every node x in addition to the main key kx also has an auxiliary key that we will denote by ax, and for these keys the heap condition is satisfied, that is
    • if y is the parent of x then ay < ax

    Thus a cartesian tree is a binary rooted ordered tree, such that each of its nodes has a pair of two keys (k, a) and three conditions described are satisfied.
    Given a set of pairs, construct a cartesian tree out of them, or detect that it is not possible.

    Input

    The first line of the input file contains an integer number N -- the number of pairs you should build cartesian tree out of (1 <= N <= 50 000). The following N lines contain two numbers each -- given pairs (ki, ai). For each pair |ki|, |ai| <= 30 000. All main keys and all auxiliary keys are different, i.e. ki != kj and ai != aj for each i != j.

    Output

    On the first line of the output file print YES if it is possible to build a cartesian tree out of given pairs or NO if it is not. If the answer is positive, on the following N lines output the tree. Let nodes be numbered from 1 to N corresponding to pairs they contain as they are given in the input file. For each node output three numbers -- its parent, its left child and its right child. If the node has no parent or no corresponding child, output 0 instead.
    The input ensure these is only one possible tree.

    Sample Input

    7
    5 4
    2 2
    3 9
    0 5
    1 3
    6 6
    4 11

    Sample Output

    YES
    2 3 6
    0 5 1
    1 0 7
    5 0 0
    2 4 0
    1 0 0
    3 0 0

    思路:裸的笛卡尔树,学习新知识,这题输入唯一,一定有解,参考博客:https://blog.csdn.net/qq_36056315/article/details/79845193
    https://blog.csdn.net/code92007/article/details/94591571
    注意不要在退栈的时候改变fa指针就行,要根据退栈完毕后left和right指针进行fa的更改,不然已经定好的顺序会乱(
    const int maxm = 5e4+10;
    
    int fa[maxm], Left[maxm], Right[maxm], N;
    
    struct Node {
        int key, value, id;
        bool operator<(const Node &node) const {
            return key < node.key;
        }
    } Nodes[maxm], s[maxm];
    
    int main() {
        scanf("%d", &N);
        for(int i = 1; i <= N; ++i) {
            scanf("%d%d", &Nodes[i].key, &Nodes[i].value);
            Nodes[i].id = i;
        }
        sort(Nodes+1, Nodes+N+1);
        int top = 0;
        bool flag = false;
        for(int i = 1; i <= N; ++i) {
            while(top && s[top].value > Nodes[i].value) {
                Left[Nodes[i].id] = s[top].id;
                top--;
            }
            fa[Nodes[i].id] = s[top].id;
            fa[Left[Nodes[i].id]] = Nodes[i].id;
            if(top)
                Right[s[top].id] = Nodes[i].id;
            s[++top] = Nodes[i];
    
        }
        printf("YES
    ");
        for(int i = 1; i <= N; ++i)
            printf("%d %d %d
    ", fa[i], Left[i], Right[i]);
        return 0;
    }
    View Code
  • 相关阅读:
    典型的 C++ 程序员成长经历
    C语言中的常用文件操作
    sscanf与正则表达式
    黑莓8700 常用快捷键
    [分享]《张靓颖个人资源汇集.九月二日更新》
    [分享]【★超级女声★】【★张靓颖★】【推荐】
    [原创]根据TreeView中节点的不同,来绑定不同的dropdownlist
    [转贴]用Vs.net制作安装程序,在安装包中自动为客户创建数据库
    [转贴]在Asp.Net中的几种事务处理的方法
    冻结DataGrid中的列的比较简单的实现方法!
  • 原文地址:https://www.cnblogs.com/GRedComeT/p/12205539.html
Copyright © 2011-2022 走看看