[滴水石穿]poj 1007-DNA Sorting 结题报告【1】

原题描述：

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 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

题目理解比较简单，从第一个字符开始，统计后面比它小的字符个数，如52341，比5小的有4个，比2小的1个，同理，3---1，4---1，总和为4+1+1+1 = 7；

计算逆序和相对比较简单，一个二维循环即可，关键是排序，我这里比较偷懒，没有涉及字符串排序，而是用一个标志数组保存每个串的逆序和，然后每次输出前遍历这个数组，求出最小逆序和的位置，进而输出，并把该逆序和调为最大，以免影响后面输出结果。

通过的关键有两点：1. N要为51，M为101，具体原因不知，理论上说50,100就可以；

2. 最大值要设的足够大，应大于所有逆序和的最大值，开始时设为65535，没有通过，后来改为655350000，就行了；当然为了保险起见，可以单独求出标志数组中最大值max，这样前面说的最大值可以设为max+1；

GCC 实现代码如下：

#include <stdio.h>

#define N 51
#define M 101


int test(char *,int );
int min(int *,int );
int main()
{
    char seq[M][N];
    int flag[M];
    int n,m,i;
    int t;


    scanf("%d %d",&n,&m);
    for(i = 0;i < m;i++){
        scanf("%s",seq[i]);
        flag[i] = test(seq[i],n);
//        printf("%d\n",flag[i]);
    }


    for(i = 0;i < m;i++){
        t = min(flag,m);
        flag[t] = 655350000;
        printf("%s\n",seq[t]);
    }
    return 0;
}


int test(char *s,int m){//
    int sum = 0;
    int i,j;
    for(i = 0;i < m - 1; i++){
        for(j = i + 1;j < m;j++){
            if(s[i] > s[j])
                sum ++;
        }
    }
    return sum;
}
int min(int *num,int m){//返回最小值的位置
    int i;
    int min = num[0],place = 0;


    for(i = 0;i < m;i++){
        if(num[i]  <  min){
            min = num[i];
            place = i;
        }
    }
    return place;

}

To myself：趁做这道题，好好复习各种排序算法，包括字符串排序。