zoukankan      html  css  js  c++  java
  • POJ 1007 DNA Sorting(sort函数的使用)

    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
     1 /*
     2 问题 输入每个字符串的长度n和字符串的个数,计算并将这些字符串的按照它的逆序数排序输出
     3 解题思路 将一个字符串和它的逆序数存入一个结构体数组中,使用sort按照逆序数排序输出即可。 
     4 */ 
     5 #include<string>
     6 #include<cstring>
     7 #include<iostream>
     8 #include<algorithm>
     9 using namespace std;
    10 
    11 struct Dstr{
    12     string str;
    13     int cou;
    14 }Dstrs[110];
    15 int unsortnum(char *str);
    16 
    17 bool cmp(struct Dstr a,struct Dstr b){
    18     return a.cou<b.cou;
    19 }
    20 
    21 int main()
    22 {
    23     int i,n,m;
    24     char temp[60];
    25     while(scanf("%d%d",&n,&m) != EOF){
    26         for(i=0;i<m;i++){
    27             scanf("%s",&temp);
    28             Dstrs[i].str=temp;
    29             Dstrs[i].cou=unsortnum(temp);
    30         }
    31         
    32         sort(Dstrs,Dstrs+m,cmp);
    33         
    34         for(i=0;i<m;i++){
    35             printf(Dstrs[i].str.c_str());
    36             printf("
    ");
    37         }
    38     }
    39     return 0;
    40 } 
    41 
    42 int unsortnum(char *str){
    43     int len=strlen(str),i,j,un=0;
    44     for(i=0;i<len-1;i++){
    45         for(j=i+1;j<len;j++){
    46             if(str[i] > str[j])    un++;
    47         }
    48     }
    49     //printf("%s %d
    ",str,un);
    50     return un;
    51 }
  • 相关阅读:
    Android 开发笔记___存储方式__共享参数__sharedprefences
    Android 开发笔记___登陆app
    Android 开发笔记___alertDialog
    Android 开发笔记___Intent的使用
    Android 开发笔记___Activity的生命周期
    Android 开发笔记___AutoComplateTextView__自动完成文本框
    单例模式
    程序设计模式六大原则--个人理解
    简单理解适配器模式
    呱呱乐
  • 原文地址:https://www.cnblogs.com/wenzhixin/p/8710165.html
Copyright © 2011-2022 走看看