UVA 409 - Excuses, Excuses!（字符串）

 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 length L (  ) and will occupy columns 1 through L 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!

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题目是说给你一些关键字，再给你几个句子，输出含有关键字最多的句子（不区分大小写），有一样多的就都输出

开始是一个字符一个字符的赋给str（句子）一直RE不明觉厉= =

于是改用gets输入str，用num[i]保存第i个句子出现的关键字数目，然后对每个字符进行处理赋给字符数组s，大写字母改成小写字母，只要不是字母就变成 ‘\0’ ，同时把s与关键字作比较，有相同的就num[i]++，找到最大值，再把所有的num与最大值比较，相同就输出第i个句子

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int key,e,t=1,nums[111];char word[111][111],str[111][111],s[111];int main(){    while(~scanf("%d%d",&key,&e))    {        printf("Excuse Set #%d\n",t++);        memset(word,0,sizeof(word));        memset(str,0,sizeof(str));        memset(nums,0,sizeof(nums));        int max=-1;        for(int i=0; i<key; i++)            scanf("%s",word[i]);        getchar();        for(int i=0; i<e; i++)        {            int j=0;            gets(str[i]);            int len=strlen(str[i]);            for(int k=0; k<len; k++)            {                if(str[i][k]>='a'&&str[i][k]<='z')                {                    s[j]=str[i][k];                    j++;                }                else if(str[i][k]>='A'&&str[i][k]<='Z')                {                    s[j]=str[i][k]+'a'-'A';                    j++;                }                else                {                    s[j]='\0';                    j=0;                    //cout<<s<<endl;                    for(int k=0; k<key; k++)                    {                        if(strcmp(s,word[k])==0)                        {                            nums[i]++;                            break;                        }                    }                }            }            //cout<<nums[i]<<endl;            if(max<nums[i])            {                max=nums[i];                //cout<<max<<endl;            }        }        for(int i=0; i<e; i++)        {            if(nums[i]==max)                printf("%s\n",str[i]);        }        printf("\n");    }    return 0;}