POJ-1007

最新推荐文章于 2021-05-19 21:53:58 发布

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最新推荐文章于 2021-05-19 21:53:58 发布

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DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 84498		Accepted: 34028

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

#include<stdio.h>
#include<string.h>

void main() 
{
    int length,line;
    int i,j,k,m,n,flag,index;
    char str[110][60];
    int len[100],len1[100];
    scanf("%d %d",&length,&line);
    for(n=0;n<line;n++) {
        scanf("%s",str[n]);
        for(i=0,m=0;i<length;i++) {
            for(j=i+1;j<length;j++){ 
		if(str[n][i]>str[n][j]) m++;		                 
            }
        }
        len[n]=m;
    }
    for(j=0,flag=0;j<line;j++) {
        for(i=0,m=60,index=0;i<line;i++){
            for(k=0;k<j;k++){ 
                if(len1[k]==i) { flag=1; break;}
                else flag=0;
            }
            if(flag==1) continue;
            else{
                if(len[i]<m){
                    m=len[i];
                    index=i;
                    len1[j]=i;
                }
            }
        }
        printf("%s\n",str[index]);
    }
}

wrong answer!