Problem B: Excuses, Excuses!

最新推荐文章于 2021-01-25 02:04:03 发布

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原文链接：http://www.cnblogs.com/NYNU-ACM/p/4236885.html

本文介绍了一个程序设计问题，旨在通过搜索一系列关键词来识别最糟糕的借口。程序接收多个数据集，每个数据集包含关键词和借口，然后找出含有最多关键词实例的借口。文章提供了输入输出示例，以及一个C++实现的代码示例。

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Problem B: Excuses, Excuses! Time Limit: 1 Sec Memory Limit: 128 MB Submit: 4 Solved: 2 [Submit][Status][Web Board] Description 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 ( tex2html_wrap_inline30 ) defines the number of keywords to be used in the search. The second number ( tex2html_wrap_inline32 ) 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 ( tex2html_wrap_inline42 ) 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 3
dog
ate
homework
canary
died
My 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 5
superhighway
crazy
thermonuclear
bedroom
war
building
I 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 #1
Can you believe my dog died after eating my canary... AND MY HOMEWORK?

Excuse Set #2
I am having a superhighway built in my bedroom.
There was a thermonuclear war!
my answer:

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cstring>
#include<cctype>
#include<string>
using namespace std;
void tolow(char s[])
{
    int tt=strlen(s);
    for(int i=0;i!=tt;i++){
        if(s[i]>='A'&&s[i]<='Z')
            s[i]+=32;
    }
}
int id(char a[],char b[])//注意调用函数中a为长串，b为短串；
{
    tolow(a);
    tolow(b);
    int str1=strlen(a);
    int str2=strlen(b);
    int k,count =0;
    for(int i=0;i!=str1;i++){
        k=0;
        for(int j=i;k!=str2;k++,j++)
        {
            if(a[j]!=b[k])
                break;
        }
        if(k>=str2){
            count++;
            break;
        }
    }
    if(count)
        return 1;
    else
        return 0;
}
int main()
{
    int n,m;
    int q=0;
    while(cin>>n>>m)
    {
        q++;
        char b[100][30];
        for(int i=0;i!=n;i++)
            cin>>b[i];
            getchar();
        char a[100][200];
        char aa[100][200];
        for(int j=0;j!=m;j++){
            gets(a[j]);
            strcpy(aa[j],a[j]);
        }
        int f=0,h=0;
        int sum[100]={0};
        for(f=0;f!=m;f++){
            for(h=0;h!=n;h++){
                sum[f]+=id(a[f],b[h]);
            }
        }
        int MAX=0;
        for (int i=0;i!=m;i++)
            if(MAX <sum[i])MAX=sum[i];
        printf("Excuse Set #%d\n",q);
        for(int j=0;j!=m;j++)
            if(MAX==sum[j])
               puts(aa[j]);
        cout<<endl;
 
 
 
    }
    return 0;
}