【基础题】对称排序

问题 : 对称排序

时间限制: 1 Sec  内存限制: 128 MB
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题目描述

In your job at Albatross Circus Management (yes, it's run by a bunch of clowns), you have just finished writing a program whose output is a list of names in nondescending order by length (so that each name is at least as long as the one preceding it). However, your boss does not like the way the output looks, and instead wants the output to appear more symmetric, with the shorter strings at the top and bottom and the longer strings in the middle. His rule is that each pair of names belongs on opposite ends of the list, and the first name in the pair is always in the top part of the list. In the first example set below, Bo and Pat are the first pair, Jean and Kevin the second pair, etc.

输入

The input consists of one or more sets of strings, followed by a final line containing only the value 0. Each set starts with a line containing an integer, n, which is the number of strings in the set, followed by n strings, one per line, NOT SORTED. None of the strings contain spaces. There is at least one and no more than 15 strings per set. Each string is at most 25 characters long.

输出

For each input set print "SET n" on a line, where n starts at 1, followed by the output set as shown in the sample output.
If length of two strings is equal,arrange them as the original order.(HINT: StableSort recommanded)

样例输入

7
Bo
Pat
Jean
Kevin
Claude
William
Marybeth
6
Jim
Ben
Zoe
Joey
Frederick
Annabelle
5
John
Bill
Fran
Stan
Cece
0

样例输出

SET 1
Bo
Jean
Claude
Marybeth
William
Kevin
Pat
SET 2
Jim
Zoe
Frederick
Annabelle
Joey
Ben
SET 3
John
Fran
Cece
Stan
Bill

题目大意：
给你几个字符串，wants the output to appear more symmetric，就是让这几句字符串输出时中间长两边短。
思路：
将给出几句字符串用stable_sort()按升序排序，然后先输出奇数的字符串，再输出偶数的字符串，如n=7，则输出
1,3,5,7,6,4,2；n=6则输出1,3,5,6,4,2。
代码如下：

 #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    typedef struct Px
    {
        char s[30];
    }Word;
    bool cmp(Word x,Word y)
    {
        int l1,l2;
        l1=strlen(x.s);
        l2=strlen(y.s);
        if(l1<l2)
        return 1;
        else return 0;
    }
    int main()
    {
        int t,x=1;
        Word a[20];
        while(scanf("%d",&t),t)
        {
            memset(a,0,sizeof(a));
            int i;
            for(i=0;i<t;i++)
             scanf("%s",a[i].s);
            stable_sort(a,a+t,cmp);
            printf("SET %d
",x);
            for(i=0;i<t;i+=2)
            {
                printf("%s
",a[i].s);
            }
            if(t&1)
            t=t-1;
            for(i=t-1;i>=1;i-=2)
            {
                printf("%s
",a[i].s);
            }
            x++;
        }
        return 0;
    }

反思：

我的思路是把奇数字符串分给一个数组，偶数的字符串分给一个数组，然后用stable_sort()给两个数组一个升序一个降序，显然我这个方法很麻烦，而且，当遇到如第三组测试数据时，每个字符串的长度相等，所以排序后偶数数组并没有变化，输出仍是Bill在前，Still在后，而上面的思路，不但效率高，而且解决了这个问题。

该题用到的C++知识点：

1、sort()排序函数的使用：

传送门：http://blog.csdn.net/zzzmmmkkk/article/details/4266888/

2、当用到字符串数组时，可以先考虑，结构体数组。

3、C/C++中strlen()详解和strlen()与sizeof()的区别：

传送门：http://blog.csdn.net/smf0504/article/details/51372351

4、C/C++中memset();的用法；

传送门：http://blog.csdn.net/binnygoal/article/details/6460887