POJ1007 DNA Sorting 解题报告

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#sorting #struct #integer #input #each #output

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本文介绍了一种基于DNA字符串的排序算法，该算法通过计算字符串中字母间的逆序对数量来评估其“有序度”，并据此对字符串进行排序。示例展示了如何使用C语言实现这一算法。

DNA Sorting

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 6

AACATGAAGG

TTTTGGCCAA

TTTGGCCAAA

GATCAGATTT

CCCGGGGGGA

ATCGATGCAT

Sample Output

CCCGGGGGGA

AACATGAAGG

GATCAGATTT

ATCGATGCAT

TTTTGGCCAA

TTTGGCCAAA

Source

East Central North America 1998

一道水题。

AC代码

#include<stdio.h> #include<stdlib.h> struct node { int a,b; }no[120]; int cmp(const void *a,const void *b) { struct node *aa=(struct node*)a; //注意这个地方 struct node *bb=(struct node*)b; if(aa->b==bb->b) return aa->a-bb->a; return aa->b-bb->b; } int main() { char s[120][60]; int m,n,i,j,k,t; scanf("%d%d",&n,&m); for(i=0;i<m;i++) { scanf("%s",s[i]); no[i].a=i; } for(i=0;i<m;i++) { t=0; for(j=0;j<n-1;j++) { for(k=j+1;k<n;k++) if(s[i][j]>s[i][k]) t++; } no[i].b=t; } qsort(no,m,sizeof(no[0]),cmp); for(i=0;i<m;i++) printf("%s/n",s[no[i].a]); return 0; }

值得一提的是结构题的快排函数的写法。

比如这道题要写成

struct node *aa=(struct node*)a;

而不能是

struct node *aa=(node*)a;

后者会导致CE，这个很奇怪啊，就POJ的编译不了，贡献了两个CE啊，ZQUOJ等其他的都可以过的。以后保守起见都要这样写了。