NYoj283对称排序

最新推荐文章于 2020-11-01 14:23:35 发布

原创最新推荐文章于 2020-11-01 14:23:35 发布 · 314 阅读

1 ·

CC 4.0 BY-SA版权

入门训练同时被 2 个专栏收录

39 篇文章

订阅专栏

其他排序

11 篇文章

订阅专栏

本文介绍了一种名为“对称排序”的算法实现方法，该算法主要用于处理字符串集合，通过先按字符串长度排序再进行特殊输出的方式，使得输出的字符串列表呈现出上下对称的特点。文章详细解释了题目要求，并提供了完整的C++代码示例。

对称排序

时间限制：1000 ms | 内存限制：65535 KB

难度：1

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

题意
很好理解
就是将字符串先按长短排个序然后按题意输出就好

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct ST{
	char s[30];
}p[20];

int cmp(ST a,ST b)
{
	int lena=strlen(a.s);
	int lenb=strlen(b.s);
	if(lena<lenb)
	return 1;
	else
	return 0;
}
int main()
{
	int n,k=0;
	while(~scanf("%d",&n),n){
		memset(p,0,sizeof(p));
		for(int i = 0;i < n;i ++)
			scanf("%s",p[i].s);
			
		printf("SET %d\n",++k);
		sort(p,p+n,cmp);
		for(int i = 0;i < n ;i+=2)
		printf("%s\n",p[i].s);
		if(n&1)
		n=n-1;
		for(int i = n-1;i >= 0 ;i-=2)
		printf("%s\n",p[i].s);
		
	}
return 0;
}