POJ1007 DNA Sorting

时间：2015-01-05 15:04:28 阅读：194 评论：0 收藏：0 [点我收藏+]

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 85026		Accepted: 34238

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

#include <stdio.h>
#include <string.h>
#include <algorithm>
using std::sort;
#define maxn 102

struct Node {
    int inverNum;
    char str[52];
} arr[maxn];
int N, M;

bool cmp(Node a, Node b) {
    if(a.inverNum == b.inverNum)
        return false;
    return a.inverNum < b.inverNum;
}

int getInver(char *str) {
    int sum = 0, i, j;
    for(i = 0; i < N - 1; ++i)
        for(j = i + 1; j < N; ++j)
            if(str[i] > str[j]) ++sum;
    return sum;
}

int main() {
    // freopen("stdin.txt", "r", stdin);
    int i, j;
    while(scanf("%d%d", &N, &M) == 2) {
        for (i = 0; i < M; ++i) {
            scanf("%s", arr[i].str);
            arr[i].inverNum = getInver(arr[i].str);
        }
        sort(arr, arr + M, cmp);
        for (i = 0; i < M; ++i)
            puts(arr[i].str);
    }
    return 0;
}

POJ1007 DNA Sorting

标签：poj1007

原文地址：http://blog.csdn.net/chang_mu/article/details/41864873

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行