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.
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
1 /* 2 问题 输入每个字符串的长度n和字符串的个数，计算并将这些字符串的按照它的逆序数排序输出 3 解题思路 将一个字符串和它的逆序数存入一个结构体数组中，使用sort按照逆序数排序输出即可。 4 */ 5 #include<string> 6 #include<cstring> 7 #include<iostream> 8 #include<algorithm> 9 using namespace std; 10 11 struct Dstr{ 12 string str; 13 int cou; 14 }Dstrs[110]; 15 int unsortnum(char *str); 16 17 bool cmp(struct Dstr a,struct Dstr b){ 18 return a.cou<b.cou; 19 } 20 21 int main() 22 { 23 int i,n,m; 24 char temp[60]; 25 while(scanf("%d%d",&n,&m) != EOF){ 26 for(i=0;i<m;i++){ 27 scanf("%s",&temp); 28 Dstrs[i].str=temp; 29 Dstrs[i].cou=unsortnum(temp); 30 } 31 32 sort(Dstrs,Dstrs+m,cmp); 33 34 for(i=0;i<m;i++){ 35 printf(Dstrs[i].str.c_str()); 36 printf("\n"); 37 } 38 } 39 return 0; 40 } 41 42 int unsortnum(char *str){ 43 int len=strlen(str),i,j,un=0; 44 for(i=0;i<len-1;i++){ 45 for(j=i+1;j<len;j++){ 46 if(str[i] > str[j]) un++; 47 } 48 } 49 //printf("%s %d\n",str,un); 50 return un; 51 }
