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欢迎参加——BestCoder周年纪念赛（高质量题目+多重奖励） |
DNA SortingTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) 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. 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 |
思路：从小到大排列。。。不好说，看题；
代码：
1 #include<stdio.h>
2 #include<algorithm>
3 using namespace std;
4 typedef struct{
5 int num;
6 char str[51];
7 }node;
8 int cmp(node a,node b){
9 return a.num<b.num;
10 }
11 int search(int n,char *a){int flot=0;
12 for(int i=0;i<n;++i){
13 for(int j=i+1;j<n;++j){
14 if(a[j]-a[i]<0)flot++;
15 }
16 }
17 return flot;
18 }
19 int main(){
20 int m,n,T;
21 node dna[110];
22 scanf("%d",&T);
23 while(T--){scanf("%d%d",&n,&m);
24 int i=0;
25 for(int i=0;i<m;++i){
26 scanf("%s",dna[i].str);
27 dna[i].num=search(n,dna[i].str);
28 }
29 sort(dna,dna+m,cmp);
30 for(int i=0;i<m;++i)printf("%s\n",dna[i].str);
31 }
32 return 0;
33 }