DNA Sorting
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 83069 | Accepted: 33428 |
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
一道简单的排序题，题目意思就是求逆序数，先求出每一段DNA序列的逆序数，然后对逆序数进行排序，然后就输出排好序之后的NDA序列；
这里进行求逆序数的有一种方法很巧妙，因为题目中只有4个字母，所以就用到了这种特殊性，我们可以得到O(n)求逆序数的方法，这就需要我们平时的多积累，和我们平时用归并排序或者用树状数组都要方便，对实际问题要进行具体分析；
下面是O(n)的求逆序数的代码；
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct node
{
char s[100];//储存DNA序列
int sum;//储存每个DNA序列的逆序数
}a[100];
bool cmp(node x,node y)//比较函数
{
return x.sum<y.sum;
}
int count_inver(char *str, int len)//求逆序数
{
int i;
int cnt = 0;
int a[4] = {0};//用一个数组个保存字母出现的次数
for(i = len - 1; i >= 0; i--) {
switch (str[i]) {
case 'A':
a[1]++;
a[2]++;
a[3]++;
break;
case 'C':
a[2]++;
a[3]++;
cnt += a[1];
break;
case 'G':
a[3]++;
cnt += a[2];
break;
case 'T':
cnt += a[3];
}
}
return cnt;
}
int main()
{
int m,n,i,j;
scanf("%d%d",&n,&m);
for(i=0;i<m;i++)
{
scanf("%s",a[i].s);
a[i].sum=count_inver(a[i].s,n);
}
sort(a,a+m,cmp);
for(j=0;j<m;j++)
printf("%s\n",a[j].s);
}
由于这道题目的数据量不算太大，直接用朴素的求逆序数也能过；
逆序数就是看这个数前面有多少个数比当前的数大，这里直接用了一个二重循环；
#include <iostream>
#include <algorithm>
using namespace std;
struct f{
int num;
char w[50];
}s[101];
bool cmp(struct f a, struct f b){
return a.num < b.num;
}
int main(){
int i, len, n, j, k;
cin>>len>>n;
for(i = 0; i < n; i++){
s[i].num = 0;
cin>>s[i].w;
for(j = 1; j < len; j++)
for(k = 0; k < j; k++)
if(s[i].w[j] < s[i].w[k])//求逆序数
s[i].num++;
}
sort(s, s+n, cmp);
for(i = 0; i < n; i++)
cout<<s[i].w<<endl;
return 0;
}