HDU 3753 Alignment of Code 字符串对其 stl

Alignment of Code

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1132    Accepted Submission(s): 270

Problem 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 begins with an integer T. The next T blocks each represents a case. Each case contains one or more lines of the code up to the end of case mark-a single '@' in a single line. All lines 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 for each case.

 

Output

For each case, 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 p_i.

 

Sample Input

1  start:  integer;   // begins   herestop:   integer; //   ends   here s:   string;   c:    char;   //  temp@

 

Sample Output

start: integer; // begins herestop:  integer; // ends   heres:     string;c:     char;    // temp

 

Author

Roman Elizarov

 

Source

2010 Northeastern European Regional Contest

 

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题目大意：让一组字符串的每一列的单词左对齐。

分析：大体分为四步：

    1.将每一句中的单词拆分开，获得单个的单词

    2.获取每一列的最大宽度，也就是最长的单词的长度

    3.获取每一行的单词的个数

    4.输出时，按照每一列的宽度设置单词的宽度，注意最后面不能有空格

因此，问题的核心就是如何将每一个单词拆分开，当然可以通过扫描并以空格为分隔符进行手工拆分，但STL还提供了更加方便的函数：istringstream(string str)，头文件为<sstream>。控制宽度时自然想到setw()，因为要左对其，还要使用setiosflags(iOS::left).

#include <bits/stdc++.h>using namespace std;const int N = 1005, M = 200;string s[N][M], line;int cw[M], cn[N]; int main(){    int r = 0, c = 0;    while(getline(cin, line))    {        stringstream ss(line);        while(ss >> s[r][c])        {            if(s[r][c].length() > cw[c])                cw[c] = s[r][c].length();            ++c;        }        cn[r++] = c;        c = 0;    }     for(int i = 0; i < r; ++i)    {        for(int j = 0; j < cn[i] - 1; ++j)            cout << left << setw(cw[j] + 1) << s[i][j];        cout << s[i][cn[i] - 1] << endl;    }    return 0;}