hdoj 1379 DNA Sorting 【结构体应用】

最新推荐文章于 2018-03-15 12:44:03 发布

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最新推荐文章于 2018-03-15 12:44:03 发布

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分类专栏：结构体

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本文介绍了一个关于DNA序列的算法问题，旨在通过计算序列中逆序对的数量来衡量其“排序性”。文章提供了完整的C++代码实现，展示了如何对多个DNA字符串进行排序，并按照从最有序到最无序的方式输出。

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DNA Sorting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2176 Accepted Submission(s): 1060

Problem 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.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

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 (1 < 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''. If two or more strings are equally sorted, list them in the same order they are in the input file.

Sample Input

1

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

定义DNA序列长度：如DAABEC

对D，有DA、DA、DB、DC4对没有按字典序排列；

对A，没有；

对B，没有；

对E，只有EC1对没有按字典序排列；

对C，没有；故共5对，长度为5。

题意：给你M个DNA序列，让你对这M个序列的长度升序排列，如果长度相等按出现先后排列。

一开始把长度全当作10了，WA了一次。。。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct rec
{
    char str[100];
    int pos, val;
}num[110];
bool cmp(rec a, rec b)
{
    if(a.val != b.val)
    return a.val < b.val;
    else
    return a.pos < b.pos;
}
int main()
{
    int n, m;
    int sum;
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        for(int i = 0; i < m; i++)
        {
            scanf("%s", num[i].str);
            sum = 0;
            for(int j = 0; j < n; j++)
            {
                for(int k = j + 1; k < n; k++)
                {
                    if(num[i].str[j] > num[i].str[k])
                    sum++;
                }
            }
            num[i].val = sum;
            num[i].pos = i;
        }
        sort(num, num+m, cmp);
        for(int i = 0; i < m; i++)
        printf("%s\n", num[i].str); 
    } 
    return 0;
}