POJ 题目1007 DNA Sorting （水题排序）

最新推荐文章于 2019-01-15 11:35:44 发布

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本文介绍了一种衡量DNA序列排序性的方法，通过计算序列中元素的逆序对数量来评估其排序程度，并提供了一个示例程序，用于对多个DNA字符串进行排序性测量并按从最有序到最无序的顺序进行排列。

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DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 82055		Accepted: 33030

Description

One measure of ``unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequence ``DAABEC'', this measure is 5, since D is greater than four letters to its right and E is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence ``AACEDGG'' has only one inversion (E and D)---it is nearly sorted---while the sequence ``ZWQM'' has 6 inversions (it is as unsorted as can be---exactly the reverse of sorted).

You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness'', from ``most sorted'' to ``least sorted''. All the strings are of the same length.

Input

The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (0 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.

Output

Output the list of input strings, arranged from ``most sorted'' to ``least sorted''. Since two strings can be equally sorted, then output them according to the orginal order.

Sample Input

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

Source

East Central North America 1998

ac代码

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct s
{
	char str[100];
	int num;
};
int cmp(const void *a,const void *b)
{
	return (*(struct s *)a).num-(*(struct s *)b).num;
}
int main()
{
	int m,n;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		int i,a[1000],j,k,v[1000];
		struct s b[1000];
		getchar();
		for(i=0;i<n;i++)
		{
			scanf("%s",b[i].str);
			b[i].num=0;
			memset(a,0,sizeof(a));
			//memset(v,0,sizeof(v));
			for(j=0;j<m;j++)
			{
				//if(!v[b[i].str[j]])
				//{
					for(k=j+1;k<m;k++)
					{
						if(b[i].str[j]>b[i].str[k])
						{
							b[i].num++;
					//		a[b[i].str[j]]++;
						}
					}
					//b[i].num+=a[b[i].str[j]];
					//v[b[i].str[j]]=1;
				//}
			}
		}
		qsort(b,n,sizeof(b[0]),cmp);
		for(i=0;i<n;i++)
			printf("%s\n",b[i].str);
	}
}