标签：poj 1007
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
分析：
问题的关键在于如何求解一个序列的无序度，最差肯定是O（n2）的，这里我们可以从归纳的角度分析问题，如果已知前n-1个字母序列的无序度，那么对于第n个字母，我们需要和前n-1个字母比较来更新，问题似乎没变，还是一样的复杂度，但是这里我们只有ACGT四个字母，我们想如何用更强的归纳来减少问题的规模，对于第n个字母，我们不需要显式的和前面n-1个字母比较，只要知道前面有几个A，几个C，几个G，几个T，我们就可以直接更新无序度了，如果为A，则新的无序度需要加上前n-1个字母的CGT个数，使用更强的归纳解决问题！
#include <iostream>
using namespace std;
int find_sortedness(char *seq,int length)
{
int numA=0,numC=0,numG=0,numT=0;
int unsort=0;
for(int i=0;i<length;i++)
{
if(seq[i]==‘A‘)
{
unsort+=(numC+numG+numT);
numA++;
}
else if(seq[i]==‘C‘)
{
unsort+=(numG+numT);
numC++;
}
else if(seq[i]==‘G‘)
{
unsort+=numT;
numG++;
}
else
{
numT++;
}
}
return unsort;
}
int main()
{
int length=0;
int nums=0;
cin>>length>>nums;
char** a=new char *[nums];
int* results=new int[nums];
int* print=new int[nums];
for(int i=0;i<nums;i++)
{
a[i]=new char[length];
for(int j=0;j<length;j++)
{
cin>>a[i][j];
}
results[i]=find_sortedness(a[i],length);
print[i]=i;
}
for(int i=0;i<nums;i++)
for(int j=nums-1;j>i;j--)
{
if(results[j]<results[j-1])
{
int tmp=results[j];
results[j]=results[j-1];
results[j-1]=tmp;
tmp=print[j];
print[j]=print[j-1];
print[j-1]=tmp;
}
}
for(int i=0;i<nums;i++)
{
for(int j=0;j<length;j++)
{
cout<<a[print[i]][j];
}
cout<<endl;
}
return 0;
}本文出自 “这与那” 博客，请务必保留此出处http://hereandthere.blog.51cto.com/7561107/1605819
标签：poj 1007
原文地址：http://hereandthere.blog.51cto.com/7561107/1605819
