poj 1007 DNA sorting

DNA Sorting

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 77160 Accepted: 30918

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

大致题意：给定长度为n,m个字符串，根据字母表顺序，来比较每个字符串的逆序数，根据逆序数从小到大输出

排好序的字符串，如果有些字符串逆序数相同，按照以前给定字符串的顺序输出。

poj的数据可能有问题，我用sort竟然也过了。

解题思路：其实就是求逆序数并进行字符串排序，关键点在于如何排序，如果逆第一时间想到sort或者qsort，

那么恭喜你打错了，题目要求按原给定顺序输出，实际就是稳定排序，可以使用冒泡或者STL的stable_sort()

进行排序，两者都是稳定排序。

看到网上介绍几种排序：引用网址：稳定排序和不稳定排序

1.稳定的：冒泡，插入，归并，基数。

2不稳定的：快速，选择，希尔，堆排序。

#include <stdio.h>#include <algorithm>using namespace std;struct Node{    char str[52];    int count;};bool mycmp(Node a, Node b){return a.count < b.count;}Node dna[102];int getRev(char s[], int n){    int i, j, cnt = 0, tmp;    for (i=0; i<n-1; ++i)    {        tmp = 0;        for (j=i+1; j<n; ++j)        {            if (s[i] > s[j])                tmp++;        }        cnt += tmp;    }    return cnt;}int main(){    int n, m, i, j;    Node t;    scanf("%d %d", &n, &m);    for (i=0; i<m; ++i)    {        scanf("%s", dna[i].str);        dna[i].count = getRev(dna[i].str, n);    }    /*// 冒泡排序也是稳定排序for (i=0; i<m; ++i)    {        for (j=0; j<m-i-1; ++j)        {            if (dna[j].count > dna[j+1].count)            {                t = dna[j];                dna[j] = dna[j+1];                dna[j+1] = t;            }        }    }*/    stable_sort(dna, dna+m, mycmp);// STL stabel_sort(first, end ,cmpfunc) 稳定排序    for (i=0; i<m; ++i)        puts(dna[i].str);    return 0;}