uva409（字符串）

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=350

409 - Excuses, Excuses!

Time limit: 3.000 seconds

Excuses, Excuses!

Judge Ito is having a problem with people subpoenaed for juryduty giving rather lame excuses in order to avoid serving. In orderto reduce the amount of time required listening to goofy excuses,Judge Ito has asked that you write a program that will search for alist 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 firstnumber ( ) defines the number of keywords to be used in thesearch. The second number ( ) defines the number of excuses in the set to besearched.
• Lines 2 through K+1 each contain exactly onekeyword.
• LinesK+2 through K+1+E each containexactly one excuse.
• All keywords in the keyword list will contain only contiguouslower case alphabetic characters of length L ( ) and will occupy columns 1 throughL in theinput line.
• All excuses can contain any upper or lower case alphanumericcharacter, a space, or any of the following punctuation marks[SPMamp".,!?&] not including thesquare 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) fromthe list.

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

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

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

SampleInput

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?

SampleOutput

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!

题意：给你一系列的关键单词，然后给你几个借口（单词），要你找出含有关键单词最多借口，如果有多个，可以按任意顺序输出。

说白了就是先给你m个单词,然后输入n句句子，求句子中有与m个单词一样的单词的个数的最大字符串（可能有多个字符串的拥有的值是一样的，都输出就好了）

#include<stdio.h>
#include<string.h>
#include<ctype.h>
char str[25][80],ss[80];  //str用于存储句子；ss用于存储句子中的单词，是时刻变化的
char word[25][25];  //用于存储关键词
int find(int m)//m表示有多少个单词
{
   int i;
   for(i = 0; i < m; i++)
  if(!strcmp(word[i],ss))  return 1;
   return 0;
}
int main()
{    
 //freopen("text.txt","r",stdin);
    int m,n,i,j,cases=0;//m表示输入的单词的个数，n表示输入的句子的个数
    int num[25],top,max;
    while(scanf("%d%d",&m,&n)==2)
 {
          printf("Excuse Set #%d\n",++cases);
          for(i=0;i<m;i++)//输入单词，其中word[0][25]存放第一个单词，word[1][25]存放第二个单词.......
                 scanf("%s",word[i]);
          getchar();  //不可丢，吸收后面的换行
          max=0;
          for(i = 0; i < n; i++)//输入句子，每输入一个句子就和前面的单词进行比较
          {
             gets(str[i]);
             num[i] = 0;//num[]表示句子中有多少个前面输入的单词
             int len = strlen(str[i]);//求字符串的长度
             top = 0;
             for(j = 0; j < len; j++)
             {
                  if(isalpha(str[i][j]))//是英文字母就保存在ss[]中，因为前面输入的word[]单词都是字母
                      ss[top++] = tolower(str[i][j]);
                 else
     {
                      if(top)   //如果s[]不空,就是说输入的字符串不是空的
                      {
                          ss[top] = '\0';  //使ss[]长度为top
                          if(find(m)) 
         num[i]++;
                           top = 0;
       }
                      else  continue;
     }
             }
             if(num[i] > max) max = num[i];//num[0]保存的是str[0][25]与单词的相同个数，num[1]保存的是str[0][25]与单词的相同个数
          }
          for(i=0;i<n;i++)//可能有多个
                 if(num[i]==max) printf("%s\n",str[i]);
          printf("\n");
       }
       return 0;
}