题目链接:

https://cn.vjudge.net/problem/UVA-10562

Professor Homer has been reported missing. We suspect that his recent research works might have had something to with this. But we really don't know much about what he was working on! The detectives tried to hack into his computer, but after hours of failed efforts they realized that the professor had been lot more intelligent than them. If only they could realize that the professor must have been absent minded they could get the clue rightaway. We at the crime lab were not at all surprised when the professor's works were found in a 3.5" floppy disk left inside the drive.

The disk contained only one text file in which the professor had drawn many trees with ASCII characters. Before we can proceed to the next level of investigation we would like to match the trees drawn with the ones that we have in our database. Now you are the computer geek -- we leave this trivial task for you. Convert professor's trees to our trees.

Professor's Trees

The first line of the input file (which you can assume comes from standard input) contains the number of trees, T (1 <= T <= 20)drawn in the file. Then you would have T trees, each ending with a single hash ('#') mark at the beginning of the line. All the trees drawn here are drawn vertically in top down fashion. The labels of each of node can be any printable character except for the symbols '-''|'' ' (space) and '#'. Every node that has children has a '|' symbol drawn just below itself. And in the next line there would be a series of '-' marks at least as long as to cover all the immediate children. The sample input section will hopefully clarify your doubts if there is any. No tree drawn here requires more than 200 lines, and none of them has more than 200 characters in one line.

Our Trees

Our trees are drawn with parenthesis and the node labels. Every subtree starts with an opening parenthesis and ends with a closing parenthesis; inside the parenthesis the node labels are listed. The sub trees are always listed from left to right. In our database each tree is written as a single string in one line, and they do not contain any character except for the node labels and the parenthesis pair. The node labels can be repeated if the original tree had such repetitions.

 /*
题意描述:
给出一棵多叉树,转化成括号表示法
解题思路:
将图存储再二维数组中,根据规则递归输出,无需建树
易错分析:
注意空树的输出
*/
#include<bits/stdc++.h> const int maxn=+;
int n;
char buf[maxn][maxn];
void dfs(int r,int c);
int main()
{
int T;
//freopen("testin.txt","r",stdin);
scanf("%d",&T);
getchar();//gets前吃掉换行符
while(T--){
memset(buf,,sizeof(maxn*maxn));
n=;
while(){
gets(buf[n]);
if(buf[n][] == '#')
break;
else
n++;
}
/*for(int i=0;i<n;i++)
puts(buf[i]);*/
printf("(");
if(n){//防止空树
for(int i=;i<strlen(buf[]);i++){
if(buf[][i] != ' '){
dfs(,i);
break;
}
}
}
printf(")\n");
}
return ;
}
//递归遍历并输出以buf[r][c]为根节点的树
void dfs(int r,int c){
printf("%c(",buf[r][c]);
if(r+ < n && buf[r+][c] == '|'){//有子树
int i=c;
while(i- >= && buf[r+][i-] == '-') i--;
while(buf[r+][i] == '-' && buf[r+][i] != '\0'){
if(buf[r+][i] != ' ')
dfs(r+,i);
i++;
}
}
printf(")");
}

UVa 10562 Undraw the Trees(递归遍历)的更多相关文章

  1. UVa 10562 Undraw the Trees 看图写树

    转载请注明: 仰望高端玩家的小清新 http://www.cnblogs.com/luruiyuan/ 题目大意: 题目传送门:UVa 10562Undraw the Trees 给定字符拼成的树,将 ...

  2. uva 10562 undraw the trees(烂题) ——yhx

    aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAABB4AAAM9CAYAAAA7ObAlAAAgAElEQVR4nOyd25GsupKGywVswAV8wA ...

  3. UVa 10562 Undraw the Trees

    题意: 将树的关系用字符串的形式给出 分析: 直接dfs搜索,第i行第j个如果是字母,判断i+1行j个是不是'|'是的话在第i+2行找第一个'-',找到后在第i+3行找字母,重复进行. 代码: #in ...

  4. UVA - 10562 Undraw the Trees(多叉树的dfs)

    题意:将多叉树转化为括号表示法. 分析:gets读取,dfs就好了.注意,样例中一行的最后一个字母后是没有空格的. #pragma comment(linker, "/STACK:10240 ...

  5. 看图写树 (Undraw the Trees UVA - 10562)

    题目描述: 原题:https://vjudge.net/problem/UVA-10562 题目思路: 递归找结点 //自己的代码测试过了,一直WA,贴上紫书的代码 AC代码 #include< ...

  6. [DFS遍历图]UVA10562 Undraw the Trees

    传送门: 1. UVA - 10562 2. Vjudge [看图写树]     将题目中给出的树改写为 括号表示法 即 (ROOT (SON1(...) (SON2(...)...(SONn(... ...

  7. UVa10562 Undraw the Trees

      注意点: 空树情况处理. >= && buf[r+][i-]=='-') i--; #include<cstdio> #include<cstring> ...

  8. (实用篇)PHP不用递归遍历目录下所有文件的代码

    <?php /** * PHP 非递归实现查询该目录下所有文件 * @param unknown $dir * @return multitype:|multitype:string */ fu ...

  9. 递归遍历XML节点属性和属性值

    public static XmlDocument FileMergedIntoXML(string strXmlPathPublic) { string strXmlPathPublic = str ...

随机推荐

  1. CSS 基础 例子 Image 高度设置

    body高度100%,三个image高度分别为auto,50%,60%,imge高度为当时body高度的百分比,auto时为原始高度 html代码如下: <!DOCTYPE html> & ...

  2. 两数据库Dblink数据抽取blob

    在目标数据库建一张临时表Create global temporary table test3 on commit preserve rows as select * from TEST1 ;在目前数 ...

  3. Android GridView 滑动条设置一直显示状态

    模拟GridView控件: <GridView android:id="@+id/picture_grid" android:layout_width="match ...

  4. 如何利用JUnit开展一个简单的单元测试(测试控制台输出是否正确)

    待测类(CreateString)如下: public class CreateString { public void createString() { //Output the following ...

  5. 背水一战 Windows 10 (48) - 控件(集合类): FlipView

    [源码下载] 背水一战 Windows 10 (48) - 控件(集合类): FlipView 作者:webabcd 介绍背水一战 Windows 10 之 控件(集合类) FlipView 示例Fl ...

  6. 19_python_反射

    一.内置函数(补充)          1.issubclass() -- 方法用于判断参数 class 是否是类型参数 classinfo 的子类.   语法格式:issubclass(class, ...

  7. 一个简单文本分类任务-EM算法-R语言

    一.问题介绍 概率分布模型中,有时只含有可观测变量,如单硬币投掷模型,对于每个测试样例,硬币最终是正面还是反面是可以观测的.而有时还含有不可观测变量,如三硬币投掷模型.问题这样描述,首先投掷硬币A,如 ...

  8. linux下时间问题---date

    显示日期 显示日期和时间

  9. Java按时间梯度实现异步回调接口

    1. 背景 在业务处理完之后,需要调用其他系统的接口,将相应的处理结果通知给对方,若是同步请求,假如调用的系统出现异常或是宕机等事件,会导致自身业务受到影响,事务会一直阻塞,数据库连接不够用等异常现象 ...

  10. css 图片文字居中

    1.单行文字居中 2.多行文字居中 3.利用background-position:center;的图片居中 4.利用display:table-cell;属性的图片居中 <!DOCTYPE h ...