DNA Sorting
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2203 Accepted Submission(s): 1075
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.
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
1 10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
Sample Output
CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA
这个题开始就没读懂，所以也不会做，后来学长讲完后才明白啥意思;
就是比如第一行数据AACATGAAGG 首先定义sum=0， A大于它后面字符的个数为0，sum+=0，第二个同样，第三个C大于它后面字符的个数为3，sum+=3。。。以此类推！
然后按每行数字从小到大输出字符串！！！
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 using namespace std; 5 struct as 6 { 7 char s[101]; 8 int m; 9 }aa[200]; 10 bool cmp(as s,as t) 11 { 12 return s.m < t.m; 13 } 14 int main() 15 { 16 int n,i,j,k,t,a,b,q; 17 scanf("%d",&n); 18 while(n--) 19 { 20 scanf("%d%d",&a,&b); 21 getchar(); 22 for(k=0;k<b;k++) 23 { 24 25 scanf("%s",aa[k].s); 26 { 27 for(t=0;t<a;t++) 28 { 29 int sum=0; 30 for(i=0;i<a-1;i++) 31 { 32 for(j=i+1;j<a;j++) 33 if(aa[t].s[i]>aa[t].s[j]) 34 sum++; 35 } 36 aa[t].m=sum;//将各组总数放在结构体中 37 } 38 39 } 40 } 41 sort(aa, aa+b, cmp);//排序结构体 42 for(i=0;i<b;i++) 43 printf("%s ",aa[i].s); 44 } 45 return 0; 46 }