STL 应用2 POJ 1007 DNA Sorting 题解 （STL完美解答）

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 63845		Accepted: 25209

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和

本题属于简单题，只要读懂题意就行了。 题意就是让你做排序，这里不多说了。想说的还是STL的两个算法在这里的应用：sort()和for_each()

sort()是STL排序算法,有两个版本，分别是： sort(RandomAccessIterator  first, RandomAccessIterator  last)，sort(RandomAccessIterator  first, RandomAccessIterator  last, Functor comp)。

for_each原型：for_each(InputIterator first, InputIterator last, Functor f)。

看看用STL写的AC源码：

#include <cstdio>
#include <algorithm>
using namespace std;
int n, len;
struct DNA
{
	char str[64];
	int val;
	DNA() : val(0) {}
	void evalue() {
		for (int i=len-1; i>=0; i--)
			for (int j=i; j>=0; j--)
				if ( str[j] > str[i] ) val++;
	}
}array[100];
bool comp(const DNA& a, const DNA& b) {	return a.val < b.val; }
void printdna(DNA& dna) {	puts(dna.str); }
int main()
{
	scanf("%d%d", &len, &n);
	for (int i=0; i<n; i++)
	{
		scanf("%s", array[i].str);
		array[i].evalue();
	}
	sort(array, array+n, comp);
	for_each(array, array+n, printdna);
return 0;	
}