poj 1007 DNA Sorting

DNA Sorting

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 94663 Accepted: 38069

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

题意是字符排序，要求是求出字符串逆序数，然后排序。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;struct point{int sum;char s[55];}p[105];bool cmp(struct point a,struct point b){return a.sum<b.sum;}int main(){int n,m,i,j,k;while(~scanf("%d %d",&n,&m)){for(i=0;i<m;i++){scanf("%s",&p[i].s);p[i].sum=0;for(j=0;j<n;j++){for(k=j+1;k<n;k++){if(p[i].s[j]>p[i].s[k]){p[i].sum++;}}}}sort(p,p+m,cmp);for(i=0;i<m;i++){printf("%s\n",p[i].s);}}return 0;}