UVA 487-3279

题目如下:

487-3279

Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable word or phrase. For example, you can call the University of Waterloo by dialing the memorable TUT-GLOP. Sometimes only part of the number is used to spell a word. When you get back to your hotel tonight you can order a pizza from Gino's by dialing 310-GINO. Another way to make a telephone number memorable is to group the digits in a memorable way. You could order your pizza from Pizza Hut by calling their ``three tens'' number 3-10-10-10.


The standard form of a telephone number is seven decimal digits with a hyphen between the third and fourth digits (e.g. 888-1200). The keypad of a phone supplies the mapping of letters to numbers, as follows:


A, B, and C map to 2

D, E, and F map to 3

G, H, and I map to 4

J, K, and L map to 5

M, N, and O map to 6

P, R, and S map to 7

T, U, and V map to 8

W, X, and Y map to 9


There is no mapping for Q or Z. Hyphens are not dialed, and can be added and removed as necessary. The standard form of TUT-GLOP is 888-4567, the standard form of 310-GINO is 310-4466, and the standard form of 3-10-10-10 is 310-1010.


Two telephone numbers are equivalent if they have the same standard form. (They dial the same number.)


Your company is compiling a directory of telephone numbers from local businesses. As part of the quality control process you want to check that no two (or more) businesses in the directory have the same telephone number.

Input
The first line of the input contains the number of datasets in the input. A blank line follows. The first line of each dataset specifies the number of telephone numbers in the directory (up to 100,000) as a positive integer alone on the line. The remaining lines list the telephone numbers in the directory, with each number alone on a line. Each telephone number consists of a string composed of decimal digits, uppercase letters (excluding Q and Z) and hyphens. Exactly seven of the characters in the string will be digits or letters. There's a blank line between datasets.

Output
Generate a line of output for each telephone number that appears more than once in any form. The line should give the telephone number in standard form, followed by a space, followed by the number of times the telephone number appears in the directory. Arrange the output lines by telephone number in ascending lexicographical order. If there are no duplicates in the input print the line:

No duplicates.

Print a blank line between datasets.

Sample Input

1

12
4873279
ITS-EASY
888-4567
3-10-10-10
888-GLOP
TUT-GLOP
967-11-11
310-GINO
F101010
888-1200
-4-8-7-3-2-7-9-
487-3279

Sample Output

310-1010 2
487-3279 4
888-4567 3


这道题难点在于很容易超时,因为号码可能高达100000个,所以必须优化算法。我是每读入一个号码,就把它转化为标准形式,这里大概是O(n^2)的复杂度(应该没法再优化了),读入完成(也就是转化完成)后,用qsort排序(效率挺高),再搜索相同的有多少个。在搜索时,我以前是对每个号码遍历其他的号码(双重循环大概是O(n^2)的复杂度),最后超时了orz。。,因为已经经过排序了,所以优化为只经一次循环,对第i个号码,检查它的下一个号码是否等于它。这样就不会超时了。

AC的代码如下:

#include
#include
#include
#include
int cmp(const void *a,const void *b)
{
    return strcmp((char*)a,(char *)b);
}
char tele[20],tele2[100000][20],trans[] = "22233344455566677778889999";


int main()
{
    int N;
    scanf("%d",&N);
    while(N--)
    {
        int ok=0;
        int n;
        scanf("%d\n",&n);

        int i;
        for(i=0; i<=n-1; i++)
        {
            scanf("%s",tele);
            int j,m=0;
            int len=strlen(tele);
            for(j=0; j<=len-1; j++)
            {
                if(isdigit(tele[j]))
                {
                    if(m==3)
                        tele2[i][m++]='-';
                    tele2[i][m++]=tele[j];
                }
                else if(isalpha(tele[j]))
                {
                    if(m==3)
                        tele2[i][m++]='-';
                    tele2[i][m++]=trans[tele[j]-'A'];
                }
            }
            tele2[i][m]='\0';
        }

        qsort(tele2,n,sizeof(tele2[0]),cmp);
        strcpy(tele2[n],"A");
        int Count=1;
        for(i=0; i<=n-1; i++)
        {
            if(strcmp(tele2[i],tele2[i+1])==0)
                Count++;
            else
            {
                if(Count>1)
                {
                    ok=1;
                    printf("%s %d\n",tele2[i],Count);
                }
                Count=1;
            }

        }
        if(ok==0)
            printf("No duplicates.\n");
        if(N!=0)
            printf("\n");
        memset(tele,'\0',sizeof(tele));
        memset(tele2,'\0',sizeof(tele2));
    }
    return 0;
}

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