poj1007 Presentation Error
通过率：39% 难度：较难
DNA Sorting
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 46899 |
|
Accepted: 18311 |
Description
One measure of``unsortedness'' in a sequence is the number of pairs of entries that are outof 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 itsright and E is greater than one letter to its right. This measure is called thenumber of inversions in the sequence. The sequence ``AACEDGG'' has only oneinversion (E and D)---it is nearly sorted---while the sequence ``ZWQM'' has 6inversions (it is as unsorted as can be---exactly the reverse of sorted).
You are responsible for cataloguing a sequence of DNA strings (sequencescontaining only the four letters A, C, G, and T). However, you want to catalogthem, 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 linecontains two integers: a positive integer n (0 < n <= 50) giving thelength of the strings; and a positive integer m (0 < m <= 100) giving thenumber of strings. These are followed by m lines, each containing a string oflength n.
Output
Output the list ofinput strings, arranged from ``most sorted'' to ``least sorted''. Since twostrings 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
这道题的题意是输入 N 个字符串，按照字符串的逆序数由最少到最大开始输出。解题思路是输入一个字符串后马上计算其逆序数，然后用一个数组 a[i] 存储逆序数，再把 a[i] 的值乘以 1000 ，再加上 i 。这样做的目的是排序的时候可以直接用 a[i] 进行快排，而不影响其对应的字符串（因为 m 最大为 100 ，乘以 1000 后排序可以直接得到排序结果，输出的时候按照 a[i]%1000 输出对应的数组序号）。即可得到结果。#include
#include
#include
char a[101][51];
int n[101];
//快排函数
int cmp(const void * a, const void * b)
{
return((*(int*)a-*(int*)b));
}
int main()
{
int i,r,c;
scanf("%d%d",&r,&c);
for(i=0;ia[i][m])
n[i]++;
n[i]=n[i]*1000+i; //逆序数x1000+原序号储存，快排时不影响原序号
}
qsort(n,c,sizeof(n[0]),cmp); //快排
for(i=0;i Presentation Error:呈现错误(结果是正确的，但是多输出空格或回车)
参考：http://zhidao.baidu.com/question/86523631.html
ps:自己写的代码比较大而慢，特百度下学习好代码，上面代码出自：
http://wenku.baidu.com/view/a96624d226fff705cc170a42.html
成绩：
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3029 11336237(3) dragoo1 172K 0MS C++ 475B 2013-03-11 09:34:47