POJ1007-DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 106816		Accepted: 42795

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.

链接

一、题意

        求一个DNA序列的逆序数，并按照逆序数从小到大顺序输出，对于逆序数相同的DNA按原顺序输出。

二、思路

        因为数据量不大，所以直接暴力求解就可以，对于排序可以使用STL提供的stable_sort函数。

三、代码

#include <iostream>
#include <sstream>
#include <iomanip>
#include <string>
#include <numeric>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <utility>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>

using namespace std;
typedef long long ll;
const int MAXN = 110;
const int MOD7 = 1000000007;
const int MOD9 = 1000000009;
const int INF = 2000000000;//0x7fffffff
const double EPS = 1e-9;
const double PI = 3.14159265358979;
const int dir_4r[] = { -1, 1, 0, 0 };
const int dir_4c[] = { 0, 0, -1, 1 };
const int dir_8r[] = { -1, -1, -1, 0, 0, 1, 1, 1 };
const int dir_8c[] = { -1, 0, 1, -1, 1, -1, 0, 1 };

struct Node {
	int val;
	char str[60];
};

Node node[MAXN];

bool operator<(const Node &a, const Node &b) {
	return a.val < b.val;
}

int main() {
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; ++i) {
		scanf("%s", node[i].str);
		node[i].val = 0;
		for (int j = 0; j < n; ++j)
			for (int k = j + 1; k < n; ++k)
				if (node[i].str[j] > node[i].str[k])
					node[i].val++;
	}

	stable_sort(node, node + m);
	for (int i = 0; i < m; ++i)
		printf("%s\n", node[i].str);

	//system("pause");
	return 0;
}