Eliminate Witches!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1124    Accepted Submission(s): 426

Problem Description
Kaname Madoka is a Magical Girl(Mahou Shoujo/Puella Magi). The duty of a Magical Girl is to eliminate Witches(Majo). Though sounds horrific, it is not a hard job for her as a powerful magical girl. 



One day Madoka is eliminating Witches as usual. This time she is facing a maze full of Witches. The maze consists of rooms, each lives exactly one Witch. And there is exactly one path from one room to another. So you see, the maze can be represented as a tree,
with rooms regarded as nodes on the tree. 



Madoka eliminates Witches according to the following rules: 

1. At first, Madoka enters the root node room of the maze. 

2. If the room Madoka enters lives a Witch, Madoka will eliminate it at once, and the Witch disappear. 

3. If the room has child node rooms with Witches, Madoka will choose the leftmost one and enter it. 

4. Madoka won't go back to the parent node room of a room X until Witches living in child node rooms of X are all eliminated. 



See the figure below for details about a sample maze. The numbers inside nodes indicate the order of elimination of the corresponding Witches, the strings below nodes are names of Witches, and the arrow shows the tracks Madoka travels: 




After finishes her task, Madoka just make a brief log like this: 

"walpurgis(charlotte(patricia,gertrud),elly,gisela)" 

which represents the tree-like maze identifying rooms by the names of Witches living in them. 



Akemi Homura, a classmate of Madoka, also a Magical Girl, is a mad fan of her. She wants to take detailed notes of everything Madoka do! Apparently the log Madoka made is hard to read, so Homura decide to make a new one of her own. 



The new log should contain the following information: 

1. The number of rooms of the maze 

2. Names of Witches in all rooms. 

3. The tracks Madoka travels. (represented by the number identifying the node) 



So the new log should be like this: 



walpurgis 

charlotte 

patricia 

gertrud 

elly 

gisela 

1 2 

2 3 

3 2 

2 4 

4 2 

2 1 

1 5 

5 1 

1 6 

6 1 



However, the maze may be very large, so Homura nees a program to help her. 
 
Input
The first line contains an integer T(T<=20), indicating the number of test cases. 

For each case there is only one string on a line, Madoka's log. 

It is guaranteed that the maze consists of at most 50000 rooms, and the names of Witches is a string consists of at most 10 lowercase characters, while the string of Madoka's log consists of at most 1000000 characters, which are lowercase characters, '(', ')'
or ','. 
 
Output
For each case, you should output the detailed log. 

The first line an integer N, the number of rooms of the maze. 

The following N lines, each line a string, the name of the Witches, in the order of elimination. 

The following 2(N-1) lines, each line two integers, the number of two nodes indicating the path Madoka passes. 

Output a blank line after each case. 
 
Sample Input
3
walpurgis(charlotte(patricia,gertrud),elly,gisela)
wuzetian
nanoha(fate(hayate))
 
Sample Output
6
walpurgis
charlotte
patricia
gertrud
elly
gisela
1 2
2 3
3 2
2 4
4 2
2 1
1 5
5 1
1 6
6 1 1
wuzetian 3
nanoha
fate
hayate
1 2
2 3
3 2
2 1
 
Source
 

网络赛模拟题,做来耍耍

思路:模拟树的先序遍历过程!

。但注意这不是二叉树!。用栈和队列实现就可以!!


代码:


#include <iostream>
#include <stdio.h>
#include <cstring>
#include <string>
#include <stack>
using namespace std; char str[1000005], ss[1000005];
char sh[105];
char ch[50005][105]; int main()
{
int i, k, len, t, ans, num;
bool tar;
scanf("%d", &t);
while(t--)
{
stack<int> ST;
scanf("%s", str);
len = strlen(str);
tar = true;
num = 0;
ans = 0;
for(i = 0; i < len; i++) //记录各个字符串
{
if(str[i]>='a' && str[i]<='z')
{
sh[ans++] = str[i];
tar = false;
}
else
{
if(tar) continue;
sh[ans] = 0;
strcpy(ch[num], sh);
num++;
ans = 0;
tar = true;
}
}
if(!tar)
{
sh[ans] = 0;
strcpy(ch[num], sh);
num++;
}
printf("%d\n", num);
for(i = 0; i < num; i++) //输出各个字符串
{
printf("%s\n", ch[i]);
}
k = 1;
for (i = 0; i < len; i++)
{
if(str[i] == '(') continue;
if(str[i] >= 'a' && str[i] <= 'z' && (i+1 >= len || str[i+1] < 'a' || str[i+1] > 'z'))
{
if(ST.size() > 0)
printf ("%d %d\n", ST.top(), k); //进栈过程输出
ST.push(k++);
}
if(str[i] == ',' || str[i] == ')') //出栈过程输出
{
printf("%d ", ST.top());
ST.pop();
printf("%d\n", ST.top());
}
}
putchar(10); //记得换行! !
}
return 0;
}


HDU-4041-Eliminate Witches! (11年北京网络赛!!)的更多相关文章

  1. HDU 4041 Eliminate Witches! (模拟题 ACM ICPC 2011亚洲北京赛区网络赛)

    HDU 4041 Eliminate Witches! (模拟题 ACM ICPC 2011 亚洲北京赛区网络赛题目) Eliminate Witches! Time Limit: 2000/1000 ...

  2. HDU 4041 Eliminate Witches! --模拟

    题意: 给一个字符串,表示一颗树,要求你把它整理出来,节点从1开始编号,还要输出树边. 解法: 模拟即可.因为由括号,所以可以递归地求,用map存对应关系,np存ind->name的映射,每进入 ...

  3. hihocoder1236(北京网络赛J):scores 分块+bitset

    北京网络赛的题- -.当时没思路,听大神们说是分块+bitset,想了一下发现确实可做,就试了一下,T了好多次终于过了 题意: 初始有n个人,每个人有五种能力值,现在有q个查询,每次查询给五个数代表查 ...

  4. 2015北京网络赛 D-The Celebration of Rabbits 动归+FWT

    2015北京网络赛 D-The Celebration of Rabbits 题意: 给定四个正整数n, m, L, R (1≤n,m,L,R≤1000). 设a为一个长度为2n+1的序列. 设f(x ...

  5. 2015北京网络赛 J Scores bitset+分块

    2015北京网络赛 J Scores 题意:50000组5维数据,50000个询问,问有多少组每一维都不大于询问的数据 思路:赛时没有思路,后来看解题报告也因为智商太低看了半天看不懂.bitset之前 ...

  6. 2015北京网络赛 Couple Trees 倍增算法

    2015北京网络赛 Couple Trees 题意:两棵树,求不同树上两个节点的最近公共祖先 思路:比赛时看过的队伍不是很多,没有仔细想.今天补题才发现有个 倍增算法,自己竟然不知道.  解法来自 q ...

  7. HDU 4759 Poker Shuffle(2013长春网络赛1001题)

    Poker Shuffle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tot ...

  8. HDU 5875 Function -2016 ICPC 大连赛区网络赛

    题目链接 网络赛的水实在太深,这场居然没出线zzz,差了一点点,看到这道题的的时候就剩半个小时了.上面是官方的题意题解,打完了才知道暴力就可以过,暴力我们当时是想出来了的,如果稍稍再优化一下估计就过了 ...

  9. HDU 4763 Theme Section (2013长春网络赛1005,KMP)

    Theme Section Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tot ...

随机推荐

  1. Visual c++ 2012 软件错误

    vs2012 未能正确加载"Visual C++ Language Manager Package"包 解决办法 如下图所示: 到官网下载更新即可. http://www.micr ...

  2. Centos + nginx + JBOSS AS 7 搭建Java web application

    最近做了一个Java的web app,一直想在Centos环境中搭建一个完整的web服务器,现在开始动手. 先说说环境: 操作系统: Centos 6.3 WEB服务器: nginx-1.2.5 Ap ...

  3. 项目管理Point

    1.项目管理流程 在设计阶段需要提交的成果物:类图设计(每个方法要有成形的SQL),页面设计,数据库设计,思路:需求分析->用户故事->客户确认(前三步是个迭代过程)->类图设计(了 ...

  4. JS模块加载器加载原理是怎么样的?

    路人一: 原理一:id即路径 原则.通常我们的入口是这样的: require( [ 'a', 'b' ], callback ) .这里的 'a'.'b' 都是 ModuleId.通过 id 和路径的 ...

  5. [BZOJ 3620] 似乎在梦中见过的样子 【KMP】

    题目链接:BZOJ - 3620 题目分析 这道题使用 KMP 做 O(n^2) 的暴力就能过. 首先,我们依次枚举字串左端点 l ,然后从这个左端点开始向后做一次 KMP. 然后我们枚举右端点 r  ...

  6. [CF Round #295 div2] C. DNA Alignment 【观察与思考0.0】

    题目链接:C. DNA Alignment 题目大意就不写了,因为叙述会比较麻烦..还是直接看英文题面吧. 题目分析 经过观察与思考,可以发现,构造的串 T 的每一个字符都与给定串 S 的每一个字符匹 ...

  7. matlab的cell数组

    matlab的cell数组 元胞数组: 元胞数组是MATLAB的一种特殊数据类型,可以将元胞数组看做一种无所不包的通用矩阵,或者叫做广义矩阵.组成元胞数组的元素可以是任何一种数据类型的常数或者常量,每 ...

  8. Occupy Cities

    hdu4606:http://acm.hdu.edu.cn/showproblem.php?pid=4606 题意:在一个二维坐标系中,有n个城市,坐标给出来了,然后有p个士兵要去占领这n个城市,但是 ...

  9. codeforces C. Painting Fence

    http://codeforces.com/contest/448/problem/C 题意:给你n宽度为1,高度为ai的木板,然后用刷子刷颜色,可以横着刷.刷着刷,问最少刷多少次可以全部刷上颜色. ...

  10. Astyle:代码格式化工具简明指南

    astyle是一个我自己常用的开放源码工具.它可以方便的将程序代码格式化成自己想要的样式而不必人工修改.本来嘛,作为高等生物应该优先去做一些智慧的事情,而不是把时间消耗在机器可以完美完成的事情上. 想 ...