uva 409 Excuses, Excuses!

Judge Ito is having a problem with people subpoenaed for jury duty giving rather lame excuses in order to avoid serving. In order to reduce the amount of time required listening to goofy excuses, Judge Ito has asked that you write a program that will search for a list of keywords in a list of excuses identifying lame excuses. Keywords can be matched in an excuse regardless of case.

Input

Input to your program will consist of multiple sets of data.

• Line 1 of each set will contain exactly two integers. The first number ( ) defines the number of keywords to be used in the search. The second number ( ) defines the number of excuses in the set to be searched.
• Lines 2 through K+1 each contain exactly one keyword.
• Lines K+2 through K+1+E each contain exactly one excuse.
• All keywords in the keyword list will contain only contiguous lower case alphabetic characters of lengthL ( ) and will occupy columns 1 throughL in the input line.
• All excuses can contain any upper or lower case alphanumeric character, a space, or any of the following punctuation marks [SPMamp".,!?&] not including the square brackets and will not exceed 70 characters in length.
• Excuses will contain at least 1 non-space character.

Output

For each input set, you are to print the worst excuse(s) from the list.

• The worst excuse(s) is/are defined as the excuse(s) which contains the largest number of incidences of keywords.
• If a keyword occurs more than once in an excuse, each occurrance is considered a separate incidence.
• A keyword ``occurs" in an excuse if and only if it exists in the string in contiguous form and is delimited by the beginning or end of the line or any non-alphabetic character or a space.

For each set of input, you are to print a single line with the number of the set immediately after the string ``Excuse Set #". (See the Sample Output). The following line(s) is/are to contain the worst excuse(s) one per line exactly as read in. If there is more than one worst excuse, you may print them in any order.

After each set of output, you should print a blank line.

Sample Input

5 3dogatehomeworkcanarydiedMy dog ate my homework.Can you believe my dog died after eating my canary... AND MY HOMEWORK?This excuse is so good that it contain 0 keywords.6 5superhighwaycrazythermonuclearbedroomwarbuildingI am having a superhighway built in my bedroom.I am actually crazy.1234567890.....,,,,,0987654321?????!!!!!!There was a thermonuclear war!I ate my dog, my canary, and my homework ... note outdated keywords?

Sample Output

Excuse Set #1Can you believe my dog died after eating my canary... AND MY HOMEWORK?Excuse Set #2I am having a superhighway built in my bedroom.There was a thermonuclear war!

题目大意：给出一些借口，再给出一些句子。输出包含最多借口的句子。

注意：有1个以上并列最多时，按输入顺序输出。

借口不能为句子中单词的子串，这样时非法的，例如：

1 2

cao

wo cao

wocao

输出：wo cao

解题思路：借用strstr,判断时要判断strstr返回的地址的后k(当前借口的长度）位或前一位是否为字母，为字母的不合法。

#include<stdio.h>#include<string.h>#include<ctype.h>int main() {int n, m;char s[20][20], s2[20][100], s3[20][100];int cnt, cnt2 = 1, excuse[20], temp;while (scanf("%d %d\n", &n , &m) != EOF) {cnt = 0;memset(s, 0, sizeof(s));             //初始化memset(s2, 0, sizeof(s2));excuse[20] = {0};temp = n;while (temp--) {gets(s[temp]);}cnt = 0;temp = m;while (temp--) {gets(s2[cnt]);int len = strlen(s2[cnt]);for (int i = 0; i < len; i++) {s3[cnt][i] = tolower(s2[cnt][i]);        //将字符全部转换为小写 }cnt++;}int flag;for (int i = 0; i < m; i++) {flag = 0;char *add = NULL;for (int j = 0;j < n; j++) {add = strstr(s3[i], s[j]);      //strstr函数，判断字符串是否处于另一个字符串中，若存在返回首地址int k = strlen(s[j]);  if (add == NULL) continue;        if (*(add - 1) >= 'a' && *(add - 1) <= 'z') continue; if (*(add + k) >= 'a' && *(add + k) <= 'z') continue; flag++;}excuse[i] = flag;}int max = 0;for (int i = 0; i < m; i++) {                 //寻找借口词最多的借口if (excuse[i] >= max) max = excuse[i];}printf("Excuse Set #%d\n", cnt2++);for (int i = 0; i < m; i++) {if (excuse[i] == max) {printf("%s\n", s2[i]);}}printf("\n");}return 0;}