[POJ 1007] DNA Sorting C++解题

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 77786		Accepted: 31201

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

中文翻译：

1007 DNA 排序

题目大意：

序列“未排序程度”的一个计算方式是元素乱序的元素对个数。例如：在单词序列“DAABEC'”中，因为D大于右边四个单词，E大于C，所以计算结果为5。这种计算方法称为序列的逆序数。序列“AACEDGG”逆序数为1（E与D）——近似排序，而序列``ZWQM'' 逆序数为6（它是已排序序列的反序）。

你的任务是分类DNA字符串（只有ACGT四个字符）。但是你分类它们的方法不是字典序，而是逆序数，排序程度从好到差。所有字符串长度相同。

输入：

第一行包含两个数：一个正整数n（0<n<=50）表示字符串长度，一个正整数m（0<m<=100）表示字符串个数。接下来m行，每行一个长度为n的字符串。

输出：

输出输入字符串列表，按排序程度从好到差。如果逆序数相同，就原来顺序输出。

样例输入：

10 6

AACATGAAGG

TTTTGGCCAA

TTTGGCCAAA

GATCAGATTT

CCCGGGGGGA

ATCGATGCAT

样例输出:

CCCGGGGGGA

AACATGAAGG

GATCAGATTT

ATCGATGCAT

TTTTGGCCAA

TTTGGCCAAA

解决思路

这是一道比较简单的排序题，我用的是选择排序。

源码

/*
poj 1000
version:1.0
author:Knight
Email:S.Knight.Work@gmail.com
*/

#include<cstdio>

using namespace std;

struct _stru_DNA

{

char String[52];

int Measure;

};

_stru_DNA DNA[110];

int n,m;

//计算第Index条DNA的Measure

void CountMeasure(int Index);

//DNA排序

void SortDNA();

int main(void)

{

int i;

scanf("%d%d", &n, &m);

for (i=0; i<m; i++)

{

scanf("%s", DNA[i].String);

CountMeasure(i);

//printf("%d\n", DNA[i].Measure);

}

SortDNA();

for (i=0; i<m; i++)

{

printf("%s\n", DNA[i].String);

//printf("%d\n", DNA[i].Measure);

}

return 0;

}

//计算第Index条DNA的Measure

void CountMeasure(int Index)

{

int i,j;

int Measure = 0;

for (i=0; i<n-1; i++)

{

if ('A' == DNA[Index].String[i])

{

continue;

}

for (j=i+1; j<n; j++)

{

if (DNA[Index].String[i] > DNA[Index].String[j])

{

Measure++;

}

}

}

DNA[Index].Measure = Measure;

}

//DNA排序

void SortDNA()

{

int i,j;

int MinIndex;

_stru_DNA Tmp;

for (i=0; i<m-1; i++)

{

MinIndex = i;

for (j=i+1; j<m; j++)

{

if (DNA[j].Measure < DNA[MinIndex].Measure)

{

MinIndex = j;

}

}

Tmp = DNA[i];

DNA[i] = DNA[MinIndex];

DNA[MinIndex] = Tmp;

}

}