Description

You are working in a team that writes Incredibly Customizable Programming Codewriter (ICPC) which is basically a text editor with bells and whistles. You are working on a module that takes a piece of code containing some definitions or other tabular information and aligns each column on a fixed vertical position, while keeping the resulting code as short as possible, making sure that only whitespaces that are absolutely required stay in the code. So, that the first words on each line are printed at position p1 = 1; the second words on each line are printed at the minimal possible position p2, such that all first words end at or before position p2 - 2; the third words on each line are printed at the minimal possible position p3, such that all second words end at or before position p3 - 2, etc. For the purpose of this problem, the code consists of multiple lines. Each line consists of one or more words separated by spaces. Each word can contain uppercase and lowercase Latin letters, all ASCII punctuation marks, separators, and other non-whitespace ASCII characters (ASCII codes 33 to 126 inclusive). Whitespace consists of space characters (ASCII code 32).

Input

The input file contains one or more lines of the code up to the end of file. All lines (including the last one) are terminated by a standard end-of-line sequence in the file. Each line contains at least one word, each word is 1 to 80 characters long (inclusive). Words are separated by one or more spaces. Lines of the code can have both leading and trailing spaces. Each line in the input file is at most 180 characters long. There are at most 1000 lines in the input file.

Output

Write to the output file the reformatted, aligned code that consists of the same number of lines, with the same words in the same order, without trailing and leading spaces, separated by one or more spaces such that i-th word on each line starts at the same position pi.

Sample Input

  start:  integer;    // begins here
stop: integer; // ends here
s: string;
c: char; // temp

Sample Output

start: integer; // begins here
stop: integer; // ends here
s: string;
c: char; // temp 代码超时!!!!!!!!!!!!!!!
#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <sstream>
using namespace std; vector<string> code[]; //1000行单词
int len[];
int main()
{
int n = ; //len中存储一行中每个单词的长度
string line;
while((getline(cin,line)) != NULL){ n++;
stringstream ss(line); //不用这个会累死的 int q = ; //q为每行单词第几个数
string word;
while(ss >> word){
int t = word.length();
if(n == )len[q] = t;
else if(len[q] < t)len[q] = t; //同一列单词中最长的
q++;
code[n].push_back(word);
}
} for(int i = ;i <= n;i++){ int N = code[i].size(); for(int j = ;j < N;j++){ line = code[i][j];
int q = line.length();
cout << code[i][j]; for( int p = ;p <= len[j] - q;p++)printf(" ");
}
putchar('\n');
} // system("pause");
return ;
}

n++的位置不对 导致插入向量的字符串存在问题

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <sstream>
using namespace std; vector<string> code[]; //1000行单词
int len[];
int main()
{
int n = ; //len中存储一行中每个单词的长度
string line;
while((getline(cin,line)) != NULL){
stringstream ss(line); //不用这个会累死的 int q = ; //q为每行单词第几个数
string word;
while(ss >> word){
int t = word.length();
if(n == )len[q] = t;
else if(len[q] < t)len[q] = t; //同一列单词中最长的
q++;
code[n].push_back(word);
}
n++;
}
for(int i = ;i < n;i++){ int N = code[i].size(); for(int j = ;j < N;j++){ line = code[i][j];
int q = line.length();
cout << code[i][j]; for( int p = ;p <= len[j] - q;p++)printf(" ");
}
putchar('\n');
} // system("pause");
return ;
}

补充几点

在vector中查找特定的元素

用 find 函数,头文件#include <algorithm>

vector<int> x;
x.push_back();
x.push_back();
x.push_back();
vector<int>::iterator iter;
iter = find(x.begin(), x.end(), );

清空vector

vector<int>student;
student.clear();

poj 3959 Alignment of Code <vector>“字符串”的更多相关文章

  1. [刷题]算法竞赛入门经典(第2版) 5-1/UVa1593 - Alignment of Code

    书上具体所有题目:http://pan.baidu.com/s/1hssH0KO 代码:(Accepted,0 ms) //UVa1593 - Alignment of Code #include&l ...

  2. UVA 1593 Alignment of Code(紫书习题5-1 字符串流)

    You are working in a team that writes Incredibly Customizable Programming Codewriter (ICPC) which is ...

  3. POJ 1836 Alignment

    Alignment Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 11450 Accepted: 3647 Descriptio ...

  4. POJ 3174 Alignment of the Planets (暴力求解)

    题意:给定 n 个坐标,问你三个共线的有多少组. 析:这个题真是坑啊,写着 n <= 770,那么一秒时间,三个循环肯定超时啊,我一直不敢写了,换了好几种方法都WA了,也不知道为什么,在比赛时坑 ...

  5. POJ 1208 The Blocks Problem --vector

    http://poj.org/problem?id=1208 晚点仔细看 https://blog.csdn.net/yxz8102/article/details/53098575 #include ...

  6. poj 3415 后缀数组 两个字符串中长度不小于 k 的公共子串的个数

    Common Substrings Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 11469   Accepted: 379 ...

  7. poj 2774 后缀数组 两个字符串的最长公共子串

    Long Long Message Time Limit: 4000MS   Memory Limit: 131072K Total Submissions: 31904   Accepted: 12 ...

  8. POJ 1836 Alignment (双向DP)

    Alignment Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 10804   Accepted: 3464 Descri ...

  9. 代码对齐 (Alignment of Code,ACM/ICPC NEERC 2010,UVa1593)

    题目描述: 解题思路: 输入时提出单个字符串,并用一个数组记录每列最长长度,格式化输出 #include <iostream> #include <algorithm> #in ...

随机推荐

  1. 练习—单链表—Swap Nodes in Pairs

    Given a linked list, swap every two adjacent nodes and return its head. For example, Given 1->2-& ...

  2. 对面向对象程序设计(OOP)的认识

    前言 本文主要介绍面向对象(OO)程序设计,以维基百科的解释: 面向对象程序设计(英语:Object-oriented programming,缩写:OOP),指一种程序设计范型,同时也是一种程序开发 ...

  3. 序列!序列!- 零基础入门学习Python016

    序列!序列! 让编程改变世界 Change the world by program 你可能发现了,小甲鱼把列表.元组和字符串放在一块儿来讲解是有道理的,我们发现Ta们之间有很多共同点: 1. 都可以 ...

  4. MVC 学习随笔(一)

    Model的绑定. (一)使用NameValueCollectionValueProvider C# 对NameValueCollectionValueProvider的支持是通过下面的类实现的 // ...

  5. php爬虫的两种思路

    写php爬虫可能最大的问题就是php脚本执行时间的问题了,对于这个问题,我找到了两种解决方法. 第一种通过代码set_time_limit(0)或者ini_set("max_executio ...

  6. Windows 8.1 with update 官方最新镜像汇总(全)

    Windows 8.1 with update 官方最新镜像汇总,发布日期: 2014/12/16,Microsoft MSDN. 镜像更新日志: 12/29:32位大客户专业版中文版12/24:64 ...

  7. 转:C# 定时任务实现

    原文地址:http://blog.csdn.net/Netself/article/details/5766398 C#实现的定时任务类,核心代码如下: 以下代码可直接封装成 TimerTask.dl ...

  8. linux下面测试网络带宽 (转载)

    利用bmon/nload/iftop/vnstat/iptraf实时查看网络带宽状况 一.添加yum源方便安装bmon# rpm -Uhv http://apt.sw.be/redhat/el5/en ...

  9. iTunes 重新提交代码步骤

    1.选择View Details 2.右侧Links-Binary Details选项 3.Reject This Binary

  10. C#使用自定义字体(从文件获取)

    在进行软件开发,尤其是开发WinForm程序时,有时为了实现界面的美化,不可避免的需要使用一些特殊的字体,但是在开发完成之后,将程序移到其他的机器上时,由于这些机器可能没有安装相应的字体,所以整个界面 ...