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.
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
題解:找出各字符串中錯誤排序次數是多少,累計錯誤排序次數是利用反序累加,利用反序,就知道當前字符與後面字符是否大小排序,若不是就累加後面小於它的字符,所有字符串都累加錯誤次序後就按照error排序從小到大。第一次的時候WA是因爲沒看到如果錯誤次數相等就按照原來順序。。。然後改了cmp函數,讓a.error>=b.error就不會出現錯誤相同順序不同!
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
using namespace std;
struct DNA{
string str;
int error;
};
int getError(string str){
int sum = 0;
int count[4] = {0,0,0,0};
int len = str.length()-1;
for(int i = len; i >=0 ;i--){
char temp = str.at(i);
switch(temp){
case 'A':count[0]++;break;
case 'C':count[1]++,sum=sum+count[0];break;
case 'G':count[2]++,sum=sum+count[1]+count[0];break;
case 'T':sum=sum+count[0]+count[1]+count[2];break;
}
}
return sum;
}
bool cmp(DNA a,DNA b){
return a.error>=b.error?false:true;
}
int main(){
int n,m;
cin>>n>>m;
DNA d[105];
for(int i = 0; i < m ;i++){
string str;cin>>str;
d[i].str = str;d[i].error = getError(str);
}
sort(d,d+m,cmp);
for(int i = 0;i < m;i++){
cout<<d[i].str<<endl;
}
return 0;
}