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POJ 1007 DNA Sorting

By Sandra Sullivan,2014-12-01 11:51
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POJ 1007 DNA Sorting

POJ 1007 DNA Sorting

简单题;暴力计算 32Ms

DNA Sorting

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 50312 Accepted: 19682

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

#include<iostream> #include<string>

#include<algorithm> using namespace std; typedef struct

{

string str;

int num; //逆序数

}Node;

int cmp(Node a, Node b) //sort 的比较函数 {

return a.num < b.num; }

int main()

{

int n, m, i, sum;

Node node[100];

while(scanf("%d", &n) != EOF) //控制输入

{

scanf("%d", &m);

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

cin >> node[i].str;

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

{

sum = 0;

for(int j = 0; j < n; j ++)

{

for(int k = j + 1; k < n; k ++) //计算逆序数

{

if(node[i].str[j] > node[i].str[k])

sum ++;

}

}

node[i].num = sum;

}

sort(node, node + m, cmp); //排序

//输出 for(i = 0; i < m; i ++)

cout << node[i].str << endl;

}

return 0;

}

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