正文
POJ 1007
小程序：扫一扫查出行【扫一扫了解最新限行尾号】
复制小程序
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
解法一：逆序数+快排（对cmp的完美诠释，原谅我刚刚学）+结构体（第一次在OJ用结构体）——
由于这道题目的数据量不算太大，直接用朴素的求逆序数绝对可以；
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>
#include <string>
using namespace std; typedef struct f{
int num;
string w;
}data;
bool cmp( data a, data b ){
return a.num < b.num;
}
int main(){
int i, len, n, j, k;
cin>>len>>n;
data *s = new data[n];
for(i = ; i < n; i++){
s[i].num = ;
cin>>s[i].w;
for(j = ; j < len; j++)
for(k = ; 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 = ; i < n; i++)
cout<<s[i].w<<endl;
return ;
}
解法二： 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；
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct node
{
char s[];//储存DNA序列
int sum;//储存每个DNA序列的逆序数
}a[];
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 = ;
int a[] = {};//用一个数组个保存字母出现的次数
for(i = len - ; i >= ; i--) {
switch (str[i]) {
case 'A':
a[]++;
a[]++;
a[]++;
break;
case 'C':
a[]++;
a[]++;
cnt += a[];
break;
case 'G':
a[]++;
cnt += a[];
break;
case 'T':
cnt += a[];
}
}
return cnt;
}
int main()
{
int m,n,i,j;
scanf("%d%d",&n,&m);
for(i=;i<m;i++)
{
scanf("%s",a[i].s);
a[i].sum=count_inver(a[i].s,n);
}
sort(a,a+m,cmp);
for(j=;j<m;j++)
printf("%s\n",a[j].s);
}
