poj 1007 逆序问题

DNA Sorting

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 62833 Accepted: 24799

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

解题关键在于计算逆序数和对逆序数的排序。

解法一：手工排序

#include <stdio.h>#include <string.h>int main(void){int temp,n,m,i,j,k,count[103]={0};char dna[103][103];scanf("%d %d",&m,&n);while (scanf(" ")||scanf("\n")) //清除中间的空格和回车;for(i=0;i<n;i++){gets(dna[i]);//取一条DNAfor(j=0;j<m;j++){//计算逆序数for(k=j+1;k<m;k++){if (dna[i][j]>dna[i][k]) count[i]++;}}count[i]=count[i]*1000+i;//合并逆序数和序号for(j=0;j<i;j++){//排序if (count[i]<count[j]) break;}if(i!=j){//如果不是最大数，则插入temp=count[i];for(k=i;k>j;k--){count[k]=count[k-1];}count[k]=temp;}}for(i=0;i<n;i++)//输出puts(dna[count[i]%1000]);return 0;}

结果：172kb    12ms

解法二：qsort排序

#include <stdio.h>#include <string.h>#include <stdlib.h>int comp(const void *a,const void *b) //定义比较函数{return *(int *)a-*(int *)b; } int main(void){int temp,n,m,i,j,k,count[103]={0};char dna[103][103];scanf("%d %d",&m,&n);while (scanf(" ")||scanf("\n")) //清除中间的空格和回车;for(i=0;i<n;i++){gets(dna[i]);//取一条DNAfor(j=0;j<m;j++){for(k=j+1;k<m;k++){if (dna[i][j]>dna[i][k]) count[i]++;}}count[i]=count[i]*1000+i;//合并逆序数和序号}qsort(count,n,sizeof(int ),comp); //qsort排序函数for(i=0;i<n;i++)//输出puts(dna[count[i]%1000]);return 0;}

结果 172kb     16ms

总结：

开始做了好多遍，但是总是runtime error或者别的错误，后来仔细看看发现原来是数组开小了的原因。。在数据量小的情况下，手排和快排的速度差不多，看不出明显差别，但是据说快排在大量数据的时候就会相比下很慢。。