POJ 1007 DNA Sorting
2015-08-17 10:10
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链接:http://poj.org/problem?id=1007
DNA Sorting
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 89319 Accepted: 35892
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.
string of length n.
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA
Source
East Central North America 1998
大意——对于一个序列的无序性可以使用相互之间无序的元素组的个数表示。问:对于给定的多个DNA字符串(长度相同,由A, C, G和T组成),从最有序到最无序进行排列,并且输出。
思路——定义一个结构体类型DNA,包含两个成员,一个str,表示DNA序列,一个count,表示逆序数。先计算出各DNA的逆序数,然后用sort函数对其按count进行从小到大排序,最后按排好序的数组进行输出即可。
复杂度分析——时间复杂度:O(m*n^2),空间复杂度:O(m*(m+n))
附上AC代码:
DNA Sorting
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 89319 Accepted: 35892
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 isgreater 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 astring 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 6AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
Sample Output
CCCGGGGGGAAACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA
Source
East Central North America 1998
大意——对于一个序列的无序性可以使用相互之间无序的元素组的个数表示。问:对于给定的多个DNA字符串(长度相同,由A, C, G和T组成),从最有序到最无序进行排列,并且输出。
思路——定义一个结构体类型DNA,包含两个成员,一个str,表示DNA序列,一个count,表示逆序数。先计算出各DNA的逆序数,然后用sort函数对其按count进行从小到大排序,最后按排好序的数组进行输出即可。
复杂度分析——时间复杂度:O(m*n^2),空间复杂度:O(m*(m+n))
附上AC代码:
#include <iostream> #include <cstdio> #include <iomanip> #include <string> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const double PI = acos(-1.0); const int MAX = 100; struct DNA { string str; int count; } dna[MAX]; bool cmp(DNA a, DNA b); int main() { ios::sync_with_stdio(false); int n, m; while (cin >> n >> m) { for (int i=0; i<m; i++) { cin >> dna[i].str; dna[i].count = 0; } for (int i=0; i<m; i++) { for (int j=0; j<n; j++) for (int k=j+1; k<n; k++) if (dna[i].str[j] > dna[i].str[k]) dna[i].count++; } sort(dna, dna+m, cmp); for (int i=0; i<m; i++) cout << dna[i].str << endl; } return 0; } bool cmp(DNA a, DNA b) { return a.count < b.count; }
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