[POJ 1007] DNA Sorting C++解题
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时间：2015年06月08日
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.
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
1 /* 2 poj 1000 3 version:1.0 4 author:Knight 5 Email:[email protected] 6 */ 7 8 #include<cstdio> 9 10 using namespace std; 11 12 struct _stru_DNA 13 14 { 15 16 char String[52]; 17 18 int Measure; 19 20 }; 21 22 _stru_DNA DNA[110]; 23 24 int n,m; 25 26 //计算第Index条DNA的Measure 27 28 void CountMeasure(int Index); 29 30 //DNA排序 31 32 void SortDNA(); 33 34 35 36 int main(void) 37 38 { 39 40 int i; 41 42 scanf("%d%d", &n, &m); 43 44 for (i=0; i<m; i++) 45 46 { 47 48 scanf("%s", DNA[i].String); 49 50 CountMeasure(i); 51 52 //printf("%d\n", DNA[i].Measure); 53 54 } 55 56 SortDNA(); 57 58 for (i=0; i<m; i++) 59 60 { 61 62 printf("%s\n", DNA[i].String); 63 64 //printf("%d\n", DNA[i].Measure); 65 66 } 67 68 return 0; 69 70 } 71 72 //计算第Index条DNA的Measure 73 74 void CountMeasure(int Index) 75 76 { 77 78 int i,j; 79 80 int Measure = 0; 81 82 for (i=0; i<n-1; i++) 83 84 { 85 86 if (‘A‘ == DNA[Index].String[i]) 87 88 { 89 90 continue; 91 92 } 93 94 for (j=i+1; j<n; j++) 95 96 { 97 98 if (DNA[Index].String[i] > DNA[Index].String[j]) 99 100 { 101 102 Measure++; 103 104 } 105 106 } 107 108 } 109 110 DNA[Index].Measure = Measure; 111 112 } 113 114 //DNA排序 115 116 void SortDNA() 117 118 { 119 120 int i,j; 121 122 int MinIndex; 123 124 _stru_DNA Tmp; 125 126 for (i=0; i<m-1; i++) 127 128 { 129 130 MinIndex = i; 131 132 for (j=i+1; j<m; j++) 133 134 { 135 136 if (DNA[j].Measure < DNA[MinIndex].Measure) 137 138 { 139 140 MinIndex = j; 141 142 } 143 144 } 145 146 Tmp = DNA[i]; 147 148 DNA[i] = DNA[MinIndex]; 149 150 DNA[MinIndex] = Tmp; 151 152 } 153 154 }
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