poj 1007 (nyoj 160) DNA Sorting
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 75164 | Accepted: 30115 |
Description
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
Output
Sample Input
10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
Sample Output
CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA
把每个序列的逆序数求出来然后根据逆序数稳定排序后升序输出,注意qsort中cmp函数的使用,当a == b时返回0则不处理两个数
#include<iostream>
#include<stdlib.h>
#include<algorithm>
#include<string>
using namespace std;
typedef struct NODE
{
int num;
string str;
}Node; int comp(const void* a,const void* b)
{
Node* x=(Node*)a;
Node* y=(Node*)b;
return (x->num)-(y->num);
}
int main()
{
int n, m;
Node array[101];
while(cin>>n>>m)
{
int i;
for(i = 0; i < m; i++)
{
Node node;
cin>>node.str;
int j;
int count = 0;
int len = node.str.length();
for(j = 0; j < len - 1; j++)
{
int k;
if(node.str[j] == 'A')
continue;
for(k = j + 1; k < len; k++)
{
if(node.str[j] > node.str[k])
count ++;
}
}
node.num = count;
array[i] = node;
}
qsort(array, m, sizeof(Node), comp);
for(i = 0; i < m; i++)
cout<<array[i].str<<endl; }
return 0;
}
poj 1007 (nyoj 160) DNA Sorting的更多相关文章
- [POJ 1007] DNA Sorting C++解题
DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 77786 Accepted: 31201 ...
- poj 1007:DNA Sorting(水题,字符串逆序数排序)
DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 80832 Accepted: 32533 Des ...
- [POJ] #1007# DNA Sorting : 桶排序
一. 题目 DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 95052 Accepted: 382 ...
- poj 1007 DNA Sorting
DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 95437 Accepted: 38399 Des ...
- DNA Sorting POJ - 1007
DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 114211 Accepted: 45704 De ...
- poj 1007 DNA sorting (qsort)
DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 95209 Accepted: 38311 Des ...
- DNA Sorting 分类: POJ 2015-06-23 20:24 9人阅读 评论(0) 收藏
DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 88690 Accepted: 35644 Descrip ...
- 算法:POJ1007 DNA sorting
这题比较简单,重点应该在如何减少循环次数. package practice; import java.io.BufferedInputStream; import java.util.Map; im ...
- [POJ1007]DNA Sorting
[POJ1007]DNA Sorting 试题描述 One measure of ``unsortedness'' in a sequence is the number of pairs of en ...
随机推荐
- java中的循环
while循环: while(循环条件){ 循环操作 } 停止while循环有两种方式:1.不再满足while后的循环条件时,循环终止: ...
- NAT,网络地址转换详解
这个技术,是一个非常成熟的技术了,但是,为了将其弄得清楚点,体系点,也为了备忘,还是有必要在这里梳理一下! NAT:Network Address Translation. 这个主要是用在网络地址(I ...
- SqlServer数据库的查询优化
建立一个web 应用,分页浏览功能必不可少.这个问题是数据库处理中十分常见的问题.经典的数据分页方法是:ADO 纪录集分页法,也就是利用ADO自带的分页功能(利用游标)来实现分页.但这种分页方法仅适用 ...
- Neutron Networking QoS
目前,Neutron有一个QoS的proposal(https://wiki.openstack.org/wiki/Neutron/QoS#Documents),但是只有Ciscso和NVP插件实现了 ...
- label 与 input
<form onsubmit="return checkform()" > <div style=" width:100%; border:0px #0 ...
- ios外包公司—北京动点软件分享:IOS工程自动打包并发布脚本实现
前言 IOS的开发过程中,当需要给测试人员发布测试包的时候,直接使用xcode来做的效率是非常低下的.尤其是当有一点小改动需要重新出包时,那简直是个折磨的人的工作.通过一番研究后,遂决定写一系列脚本, ...
- python asyncio笔记
1.什么是coroutine coroutine,最早我是在lua里面看到的,coroutine最大的好处是可以保存堆栈,让程序得以继续执行,在python里面,一般是利用yield来实现,具体可以看 ...
- nand flash,nor flash,spi flash,片上RAM,片外RAM
Flash有掉电数据保存的特点,RAM掉电则数据丢失,但是RAM的速度更高,擦写次数理论上没有限制,而Flash则不行. Nand Flash相比其余的几种flash优势在于可擦写次数多,擦写速度快, ...
- 编译在arm板上使用的sqlite3的静动态库
采用的是sqlite-autoconf-3080002.tar.gz 解压 tar xvf sqlite-autoconf-3080002.tar.gz 进入 cd sqlite-autoconf-3 ...
- 86. Partition List
Given a linked list and a value x, partition it such that all nodes less than x come before nodes gr ...