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
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 sequenceDAABEC”, 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 sequenceZWQM” 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'', frommost 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'' toleast 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
大致的意思就是对DNA的逆序数排序
#include <iostream>
#include <string>
#include <algorithm>
#define RR freopen("input.txt","r",stdin)
#define WW freopen("ouput.txt","w",stdout)
using namespace std;
const int INF=0x3f3f3f3f;
int n,m;
struct DNA
{
string str;
int num ;
void NUM()
{
num=0;
for(int i=0;i<n;i++)
{
int sum=0;
for(int j=i-1;j>=0;j--)
{
if(str[i]<str[j])
{
sum++;
}
}
num+=sum;
}
}
}D[110];
bool cmp(DNA a,DNA b)
{
return a.num<b.num;
}
int main()
{
cin>>n>>m;
for(int i=0;i<m;i++)
{
cin>>D[i].str;
D[i].NUM();
}
sort(D,D+m,cmp);
for(int i=0;i<m;i++)
{
cout<<D[i].str<<endl;
}
return 0;
}
版权声明:本文为博主原创文章,未经博主允许不得转载。
DNA Sorting 分类: POJ 2015-06-23 20:24 9人阅读 评论(0) 收藏的更多相关文章
- hadoop调优之一:概述 分类: A1_HADOOP B3_LINUX 2015-03-13 20:51 395人阅读 评论(0) 收藏
hadoop集群性能低下的常见原因 (一)硬件环境 1.CPU/内存不足,或未充分利用 2.网络原因 3.磁盘原因 (二)map任务原因 1.输入文件中小文件过多,导致多次启动和停止JVM进程.可以设 ...
- 哈希-Snowflake Snow Snowflakes 分类: POJ 哈希 2015-08-06 20:53 2人阅读 评论(0) 收藏
Snowflake Snow Snowflakes Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 34762 Accepted: ...
- Poj 2559 最大矩形面积 v单调栈 分类: Brush Mode 2014-11-13 20:48 81人阅读 评论(0) 收藏
#include<iostream> #include<stack> #include<stdio.h> using namespace std; struct n ...
- Dijkstra with priority queue 分类: ACM TYPE 2015-07-23 20:12 4人阅读 评论(0) 收藏
POJ 1511 Invitation Cards(单源最短路,优先队列优化的Dijkstra) //================================================= ...
- NYOJ 119 士兵杀敌(三)【ST算法】 分类: Brush Mode 2014-11-13 20:56 101人阅读 评论(0) 收藏
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=119 解题思路: RMQ算法. 不会的可以去看看我总结的RMQ算法. http://blo ...
- bzoj 1041 圆上的整点 分类: Brush Mode 2014-11-11 20:15 80人阅读 评论(0) 收藏
这里先只考虑x,y都大于0的情况 如果x^2+y^2=r^2,则(r-x)(r+x)=y*y 令d=gcd(r-x,r+x),r-x=d*u^2,r+x=d*v^2,显然有gcd(u,v)=1且u&l ...
- HTTP 错误 500.19- Internal Server Error 错误解决方法 分类: Windows服务器配置 2015-01-08 20:16 131人阅读 评论(0) 收藏
1.第一种情况如下: 解决方法如下: 经过检查发现是由于先安装Framework组件,后安装iis的缘故,只需重新注册下Framework就可以了,具体步骤如下 1 打开运行,输入cmd进入到命令提示 ...
- winform Execl数据 导入到数据库(SQL) 分类: WinForm C# 2014-05-09 20:52 191人阅读 评论(0) 收藏
首先,看一下我的窗体设计: 要插入的Excel表: 编码 名称 联系人 电话 省市 备注 100 100线 张三 12345678910 北京 测试 101 101线 张三 12345678910 上 ...
- House Robber 分类: leetcode 算法 2015-07-09 20:53 2人阅读 评论(0) 收藏
DP 对于第i个状态(房子),有两种选择:偷(rob).不偷(not rob) 递推公式为: f(i)=max⎧⎩⎨⎪⎪{f(i−1)+vali,f(i−2)+vali,robi−1==0robi−1 ...
随机推荐
- 2-sat 输出任意一组可行解&拓扑排序+缩点 poj3683
Priest John's Busiest Day Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8170 Accept ...
- zjuoj 3600 Taxi Fare
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3600 Taxi Fare Time Limit: 2 Seconds ...
- vs2010打包系统必备选择.net framework 3.5sp1编译错误的解决方法
利用visual studio 2010进行打包程序,默认安装的是Framework 4.0,如果需要将3.5sp1打包到系统中一起安装(选择了"从与我的应用程序相同的位置下载系统必备组件& ...
- 用xml添加动画
在res文件夹下新建anim文件夹 在anim文件夹新建anim.xml anim.xml: <?xml version="1.0" encoding="utf-8 ...
- 异常:Struts:org.springframework.beans.factory.CannotLoadBeanClassException: Cannot find BasicDataSource
org.springframework.beans.factory.CannotLoadBeanClassException: Cannot find class [org.apache.common ...
- Uploadify在MVC中使用方法案例(上传单张图片)
在View视图中: <link href="/Scripts/uploadify-v3.2.1/uploadify.css" rel="stylesheet&quo ...
- Power Gating的设计(概述)
Leakage power随着CMOS电路工艺进程,功耗越来越大. Power Domain的开关一般通过硬件中的timer和系统层次的功耗管理软件来进行控制,需要在一下几方面做trade-off: ...
- 「thunar」给thunar增加搜索文件功能
1.安装catfish sudo apt-get install catfish 2.配置thunar,添加[自定义动作] 打开 Thunar 后,点击 Edit -> Configure cu ...
- SQL Server数据库的三种恢复模式:简单恢复模式、完整恢复模式和大容量日志恢复模式(转载)
SQL Server数据库有三种恢复模式:简单恢复模式.完整恢复模式和大容量日志恢复模式: 1.Simple 简单恢复模式, Simple模式的旧称叫”Checkpoint with truncate ...
- jQuery触发<a>标签的点击事件后URL不跳转的解决办法
有HTML代码如下: <a id="workFrame" href="pages/work.html" target="FrameBox&quo ...