ACM pku 1007　解题报告（给像我一样的新手们的忠告）

DNA Sorting
Time Limit:1000MS Memory Limit:10000K
Total Submit:6862 Accepted:2809
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

Source
East Central North America 1998

PKU 1007 GCC代码如下：

main()
{
int n,m,i,j,t,minIndex;
int rank[150]; /*存放逆序数*/
char dna[150][100]; /*存放字符串*/
scanf("%d%d",&n,&m);
/*输入字符串*/
for( i = 0 ; i < m ; i++ )
{
scanf("%s",dna[i]);
}
/*求逆序数*/
for(i = 0; i < m ; i++)
{
rank[i] = 0;
for( j=0 ; j<n-1 ; j++)
{
for(t=j+1;t<n;t++) /*这里可以优化,如果DNA[i][j]是A的话,就不要这个FOR循环了,因为没有比A再小的字母了*/
if(dna[i][t] < dna[i][j] )rank[i]++;

}/*j*/
}/*i*/

for(i=0;i<m;i++){
minIndex=i; /* MININDEX为带有最小值的数的下标*/
for(j=0;j<m;j++){ /*这个FOR将最小值的下标给MININDEX*/
if(rank[minIndex]>rank[j])minIndex=j;
}
printf("%s/n",dna[minIndex]); /*将目前最小逆序字符串输出*/
rank[minIndex]=1000;
}/*i*/
}/*main*/

由本题得出的解题忠告:把数组下标稍微设大一点.原来我设的是
int rank[101];
char dna[100][50];

心想,够了吧.那小得搞到凌晨5:00都过不了.今天把下标一改,原来写的几个版本都过了.我想杀了了!

感谢我自己的执着:)

C中字符的比较直接用 == > <....就可以了

下面是另一个通过了的版本

main()
{
int n,m,i,j,t,s;
int rank[150][2];
char dna[150][150];
scanf("%d%d",&n,&m);

for( i = 0 ; i < m ; i++ )
{
scanf("%s",dna[i]);
}
for(i = 0; i < m ; i++)
{
rank[i][0] = 0;
rank[i][1] = i;
for( j=0 ; j<n-1 ; j++)
{

if( dna[i][j] != 'A' )
for(t=j+1;t<n;t++)
if(dna[i][t] < dna[i][j] )rank[i][0]++;

}/*j*/
}/*i*/

for(i=0;i<m;i++){

for(j=i+1;j<m;j++){

if(rank[i][0]>rank[j][0]){

t = rank[i][0];
s = rank[i][1];
rank[i][0] = rank[j][0];
rank[i][1] = rank[j][1];
rank[j][0] = t;
rank[j][1] = s;
}
}

}
for(i=0;i<m;i++)printf("%s/n",dna[rank[i][1]]);

}/*main*/