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. 关于Jquery.validate.js中动态删除验证remove方法的Bug

    利用Jquery.validate.js 来做动态验证的时候,需要特定的情况下,删除添加opAmount的必须入力的Check $("#form").validate({ rule ...

  2. windows应用中调用DLL一步步试验

    试验环境: PC:win10 build 10143 IDE: vs2015 RC WinPhone: win10 build 10136 简单界面,点按钮,算加法 一.主程用C++ 1.新建visu ...

  3. java 编码转换

    在网络中爬取到的数据,编码可能与当前编译器的编码不相同,而导致可能产生显示乱码的问题.那么如何将网络的编码,转换为当前编译器认可的编码(一般为UTF-8),就是个问题了. 主要使用了两个方法: Str ...

  4. JDK Linux环境配置

    ① $sudo vi /etc/profile ② 在末尾行添加 #set java environment JAVA_HOME=/usr/local/jdk1.7.0 CLASSPATH=.:$JA ...

  5. SpringMVC整合极光推送报错ClassNotFound

    问题: 今天在做后台和极光整合的过程中,将极光部分代码整合到Dao层,在启动项目的过程中总是报错,classNotFund cn/jpush/api/push/xxxx 极光官方文档: http:// ...

  6. JAVA实现实用的ZIP压缩与解压

    http://blog.csdn.net/z69183787/article/details/38555913

  7. KEIL的宏汇编器A51介绍

    A51是一种具有通用特性和用法的重定位宏汇编器.它与Intel公司的MASM51宏汇编器具有很好兼容性,支持模块化编程,可以方便地与高级语言接口.A51宏汇编器支持汇编伪指令.宏处理指令以及汇编控制命 ...

  8. [Android]通过setImageURI设置网络上面的图片

    设置imageView显示网络上的图片 picUrl = new URL(getIntent().getExtras().getString("map_url")); Bitmap ...

  9. windows phone因为墓碑化导致“正在恢复”的分析

    我们在平时的WP使用过程中,会遇到一个问题 应用在切出,切回后, 有时候,会显示"正在恢复",并等待时间较长,才能回到用户切出时候的画面,但是这种情况并非常见,偶尔发生 有时候,直 ...

  10. 01_docker学习总结

    01 docker学习总结 toolbox https://hub.docker.com/ https://docs.docker.com/engine/installation/mac/#from- ...