poj1007

DNA Sorting

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 64183 Accepted: 25343

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 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

poj的题读起来很费劲儿，对我这种英语不好的来说更是如此。

题目中先给出一个定义unsortedness，来定义一串字符的无序程度，值越大越无序（按照字母表顺序）。然后最后各个串按照unsortedness值从大到小输出。

#include <stdio.h>#include <string.h>int main(){int n, m, sorted[101], completed[101], i, j, k, min;char str[101][51];memset(sorted, 0, sizeof(sorted) );memset(completed, 0, sizeof(completed) );scanf("%d %d", &m, &n);getchar();i=0;while(i<=n-1){scanf("%s",str[i]);i++;}for(i=0;i<=n-1;i++){for(j=0;j<=m-1;j++){for(k=j;k<=m-1;k++){if(str[i][j]>str[i][k])sorted[i]++;}}}for(i=0;i<=n-1;i++){min=i;for(j=0;j<=n-1;j++){if(sorted[j]<sorted[min])min=j;}completed[i]=min;sorted[min]=10000;}for(i=0;i<=n-1;i++){puts(str[completed[i]]);}return 0;}