POJ 1007 DNA Sorting 多个数组代替结构体

DNA Sorting

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 327  Solved: 140
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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

HINT

poj1007

Source

East Central North America 1998
        该题为我们提供了一种可代替结构体的方法（但是相比于结构体很是笨重）。题目要求我们记录一个主体及它的一些属性（总称为 块），当其中一项改变时，整个块也要发生改变，很显然，结构体可以轻易地做到这些，but，我们有两个数组同样可以实现，只是改变时，要改变两个数组。以题目中按最大值顺序排列举例，如果用结构体来做，可以用一个参数在一个步骤里实现，而用数组的话，相当于分两次改变两个数组的内容。AC代码：

#include<stdio.h>
#include<string.h>
int main()
{
int i,p,j,n,m,k,y,x;
int num[102]={0};
char mid[54],fir[102][52];
scanf("%d %d",&m,&n);
getchar();
for(i=1;i<=n;i++)
{
y=0;
gets(fir[i]);
for(j=0;j<=m-2;j++)
for(p=j+1;p<=m-1;p++)
if(fir[i][j]>fir[i][p])
y++;
num[i]=y;
}
for(i=1;i<=n;i++)
{
k=i;
for(j=i+1;j<=n;j++)
{
if(num[k]>num[j])
k=j;
}
strcpy(mid,fir[k]);
strcpy(fir[k],fir[i]);
strcpy(fir[i],mid);
x=num[k];
num[k]=num[i];
num[i]=x;
}
for(i=1;i<=n;i++)
puts(fir[i]);
return 0;
}

下面这个代码在POJ上WA了，不知道为什么。。。。

#include<stdio.h>
#include<string.h>
char sec[1500][5001];
int main()
{
int i,p,j,n,m,x,y;
char first[105][52];
char c;
scanf("%d %d",&n,&m);
getchar();
y=0;
for(i=1;i<=m;i++)
gets(first[i]);
for(i=1;i<=m;i++)
{
x=0;
for(j=0;j<=n-1;j++)
{
for(p=j+1;p<=n-1;p++)
if(first[i][j]>first[i][p])
x++;
}
if(sec[x][0]!=0)
{
first[i]
='\n';
first[i][n+1]=0;
}
strcat(sec[x],first[i]);
}
printf("%d\n",x);
for(i=0;i<=1500;i++)
if(sec[i][0]!=0)
puts(sec[i]);
return 0;
}