题目链接:https://vjudge.net/problem/UVA-1525

题目链接:https://vjudge.net/problem/POJ-1577

题目大意

  略。

分析

  建树,然后先序遍历。

代码如下

 #include <cmath>
#include <ctime>
#include <iostream>
#include <string>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <cctype>
#include <stack>
#include <deque>
#include <list>
#include <sstream>
#include <cassert>
using namespace std; #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (int)(n); ++i)
#define For(i,s,t) for (int i = (int)(s); i <= (int)(t); ++i)
#define rFor(i,t,s) for (int i = (int)(t); i >= (int)(s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) #define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl #define LOWBIT(x) ((x)&(-x)) #define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
#define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c
#define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
#define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper); #define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,0x3f,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a)) #define MP make_pair
#define PB push_back
#define ft first
#define sd second template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
in >> p.first >> p.second;
return in;
} template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
for (auto &x: v)
in >> x;
return in;
} template<typename T>
ostream &operator<<(ostream &out, vector<T> &v) {
Rep(i, v.size()) out << v[i] << " \n"[i == v.size()];
return out;
} template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
out << "[" << p.first << ", " << p.second << "]" << "\n";
return out;
} inline int gc(){
static const int BUF = 1e7;
static char buf[BUF], *bg = buf + BUF, *ed = bg; if(bg == ed) fread(bg = buf, , BUF, stdin);
return *bg++;
} inline int ri(){
int x = , f = , c = gc();
for(; c<||c>; f = c=='-'?-:f, c=gc());
for(; c>&&c<; x = x* + c - , c=gc());
return x*f;
} template<class T>
inline string toString(T x) {
ostringstream sout;
sout << x;
return sout.str();
} inline int toInt(string s) {
int v;
istringstream sin(s);
sin >> v;
return v;
} //min <= aim <= max
template<typename T>
inline bool BETWEEN(const T aim, const T min, const T max) {
return min <= aim && aim <= max;
} typedef long long LL;
typedef unsigned long long uLL;
typedef vector< int > VI;
typedef vector< bool > VB;
typedef vector< char > VC;
typedef vector< double > VD;
typedef vector< string > VS;
typedef vector< LL > VL;
typedef vector< VI > VVI;
typedef vector< VB > VVB;
typedef vector< VS > VVS;
typedef vector< VL > VVL;
typedef vector< VVI > VVVI;
typedef vector< VVL > VVVL;
typedef pair< int, int > PII;
typedef pair< LL, LL > PLL;
typedef pair< int, string > PIS;
typedef pair< string, int > PSI;
typedef pair< string, string > PSS;
typedef pair< double, double > PDD;
typedef vector< PII > VPII;
typedef vector< PLL > VPLL;
typedef vector< VPII > VVPII;
typedef vector< VPLL > VVPLL;
typedef vector< VS > VVS;
typedef map< int, int > MII;
//typedef unordered_map< int, int > uMII;
typedef map< LL, LL > MLL;
typedef map< string, int > MSI;
typedef map< int, string > MIS;
typedef set< int > SI;
typedef stack< int > SKI;
typedef queue< int > QI;
typedef priority_queue< int > PQIMax;
typedef priority_queue< int, VI, greater< int > > PQIMin;
const double EPS = 1e-;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + ;
const int maxN = 1e3 + ;
const LL ONE = ;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555; struct BSTNode {
char var;
int lc, rc, cnt; BSTNode() {}
BSTNode(char x) {
lc = rc = ;
var = x;
cnt = ;
}
}; int N;
VS leaf;
string tmp;
vector< BSTNode > bst; void insertBSTNode(char x, int rt) {
if(x < bst[rt].var) {
if(bst[rt].lc) insertBSTNode(x, bst[rt].lc);
else {
bst[rt].lc = bst.size();
bst.PB(BSTNode(x));
}
}
else if(x > bst[rt].var) {
if(bst[rt].rc) insertBSTNode(x, bst[rt].rc);
else {
bst[rt].rc = bst.size();
bst.PB(BSTNode(x));
}
}
//else ++bst[rt].cnt;
} void preOrder(int rt) {
cout << bst[rt].var;
if(bst[rt].lc) preOrder(bst[rt].lc);
if(bst[rt].rc) preOrder(bst[rt].rc);
} int main(){
//freopen("MyOutput.txt","w",stdout);
//freopen("input.txt","r",stdin);
INIT();
while(tmp != "$") {
leaf.clear();
bst.clear();
bst.PB(BSTNode()); // 0 号节点设为空节点 while(cin >> tmp && tmp != "*" && tmp != "$") leaf.PB(tmp);
bst.PB(BSTNode(leaf.back()[]));
rFor(i, leaf.size() - , ) Rep(j, leaf[i].size()) insertBSTNode(leaf[i][j], ); preOrder();
cout << endl;
}
return ;
}

UVA 1525 Falling Leaves的更多相关文章

  1. UVA - 699The Falling Leaves(递归先序二叉树)

    The Falling Leaves Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Sub ...

  2. Uva 699The Falling Leaves

    0.唔.这道题 首先要明确根节点在哪儿 初始化成pos=maxn/2; 1.因为是先序的输入方法,所以这个建树的方法很重要 void build(int p) { int v; cin>> ...

  3. UVA.699 The Falling Leaves (二叉树 思维题)

    UVA.699 The Falling Leaves (二叉树 思维题) 题意分析 理解题意花了好半天,其实就是求建完树后再一条竖线上的所有节点的权值之和,如果按照普通的建树然后在计算的方法,是不方便 ...

  4. UVa 699 The Falling Leaves(递归建树)

    UVa 699 The Falling Leaves(递归建树) 假设一棵二叉树也会落叶  而且叶子只会垂直下落   每个节点保存的值为那个节点上的叶子数   求所有叶子全部下落后   地面从左到右每 ...

  5. UVA 699 The Falling Leaves (二叉树水题)

    本文纯属原创.转载请注明出处,谢谢. http://blog.csdn.net/zip_fan. Description Each year, fall in the North Central re ...

  6. UVa 699 The Falling Leaves (树水题)

    Each year, fall in the North Central region is accompanied by the brilliant colors of the leaves on ...

  7. The Falling Leaves(建树方法)

    uva 699 紫书P159 Each year, fall in the North Central region is accompanied by the brilliant colors of ...

  8. H - The Falling Leaves

    Description Each year, fall in the North Central region is accompanied by the brilliant colors of th ...

  9. UVa699 The Falling Leaves

      // UVa699 The Falling Leaves // 题意:给一棵二叉树,每个节点都有一个水平位置:左儿子在它左边1个单位,右儿子在右边1个单位.从左向右输出每个水平位置的所有结点的权值 ...

随机推荐

  1. c/c++ int 范围的原因

    在C语言中, signed char 类型的范围为-128~127,每本教科书上也这么写,但是没有哪一本书上(包括老师)也不会给你为什么是-128~127,这个问题貌似看起来也很简单容易, 以至于不用 ...

  2. seleniumIDE command命令

    语法组成要素:command.target.value. command命令 三大类型:(action.Accessor.assertion)   操作  存储  断言 操作类型——Action 浏览 ...

  3. 一步一步学Vue(六)https://www.cnblogs.com/Johnzhang/p/7242640.html

    一步一步学Vue(六):https://www.cnblogs.com/Johnzhang/p/7237065.html  路由 一步一步学Vue(七):https://www.cnblogs.com ...

  4. FrameWork内核解析之Handler消息机制(二)

    阿里P7Android高级架构进阶视频(内含Handler视频讲解)免费学习请点击:https://space.bilibili.com/474380680 一.Handler 在Android开发的 ...

  5. CVE-2010-4258漏洞分析

    Nelson Elhage最近发现了一个内核设计上的漏洞, 通过利用这个漏洞可以将一些以前只能dos的漏洞变成可以权限提升的漏洞. 当fork一个进程在的时候, copy_process执行如下操作: ...

  6. LeetCode Array Easy 26.Remove Duplicates from Sorted Array 解答及疑惑

    Description Given a sorted array nums, remove the duplicates in-place such that each element appear ...

  7. junit单元测试报错Failed to load ApplicationContext,但是项目发布到tomcat浏览器访问没问题

    junit单元测试报错Failed to load ApplicationContext,但是项目发布到tomcat浏览器访问没问题,说明代码是没问题的,配置也没问题.开始时怀疑是我使用junit版本 ...

  8. (好题)2017-2018 ACM-ICPC, Asia Tsukuba Regional Contest F Pizza Delivery

    题意:给n个点m条边的有向图.每次使一条边反向,问你1到2的最短路变短,变长,还是不变. 解法:遇到这种题容易想到正向求一遍最短路d1,反向再求一遍最短路d2.纪录原图上的最短路为ans,然后分开考虑 ...

  9. ubuntu:beyond compare 4 This license key has been revoked 解决办法

    错误如图所示: 解决办法: (1)先用find命令找到bcompare所在位置:sudo find /home/ -name '*bcompare' ()进入 /home/whf/.config,删除 ...

  10. vue filters 日期

    <!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title> ...