POJ水题1007DNA Sorting

DNA Sorting
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 91275 Accepted: 36661
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 sequenceDAABEC”, 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 sequenceZWQM” 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'', frommost 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 frommost 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
Source
East Central North America 1998

数据结构：每个串跟一个逆序数关联，用结构体就行。
算法：数据规模不大，暴力模拟即可。
用sort按逆序数排序，因为sort是基于快排的，排序过程主要的就是交换实现，然而如果两个数相同，不会交换两个数的顺序，因此如果某两个串的逆序数相同，它们在结果中当然按输入顺序

//244K  16MS#include<cstdio>#include<cmath>#include<iostream>#include<algorithm>using namespace std;int n,m;struct _DNA{int InverNum;int Order;    char str[55];}DNA[105]; int GetInver(char s[]){    int sum=0;    for(int i=0;i<n;i++){        for(int j=i;j<n;j++){            if(s[j]<s[i])sum++;        }    }    return sum;}bool cmp(_DNA a,_DNA b){    return a.InverNum==b.InverNum? a.Order<b.Order:a.InverNum<b.InverNum;}int main(){    cin>>n>>m;    for(int i=0;i<m;i++){        scanf("%s",&DNA[i].str);        DNA[i].InverNum=GetInver(DNA[i].str); DNA[i].Order=i;    }    sort(DNA,DNA+m,cmp);    for(int i=0;i<m;i++){        printf("%s\n",DNA[i].str);    }    return 0;}