HDU1379-逆序数排序

DNA Sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1716    Accepted Submission(s): 823

Problem 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. 
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.

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 (1 < 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''. If two or more strings are equally sorted, list them in the same order they are in the input file. 

Sample Input

110 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

Source

East Central North America 1998

思路：

题目意思是：根据DNA串的逆序数时行排序，如果DNA串的逆序数相等，则保持DNA原本的相对位置不变。

所以，重点是求出字符串的逆序数，求逆序数可以用蛮力法-O(n^2)，也可以用归并排序的方法-O(nlgn)。打开链接

#include <cstdio>#include <string>#include <iostream>#include <algorithm>using namespace std;char arr[120],tmp[120];int sum = 0;struct node{int id;int inversion;char str[120];} dna[120];bool cmp(const node& x,const node& y){if (x.inversion != y.inversion)return x.inversion < y.inversion;elsereturn x.id < y.id;}void Merge(int left,int mid,int right){int i=left, j=mid+1, k=left;while(i <= mid && j <= right){if(arr[i] <= arr[j])tmp[k++]=arr[i++];else{tmp[k++]=arr[j++];sum += mid-i+1;//求逆序数}}while(i <= mid)tmp[k++]=arr[i++];while(j <= right)tmp[k++]=arr[j++];for(int i=left;i <= right;++i)arr[i]=tmp[i];}void MergeSort(int left,int right){if(left < right){int mid = (left + right) >> 1;MergeSort(left,mid);MergeSort(mid+1,right);Merge(left,mid,right);}}int main(){#ifndef ONLINE_JUDGEfreopen("2.txt","r",stdin);#endifint Case,n,m,i;scanf("%d ",&Case);while(Case--){scanf("%d%d ",&n,&m);for (i=0; i < m; ++i){scanf("%s ",arr);strcpy(dna[i].str,arr);sum = 0;MergeSort(0,n-1);dna[i].id = i;dna[i].inversion = sum;}sort(dna,dna+m,cmp);for(i=0; i < m; ++i)puts(dna[i].str);}    return 0;}