哈哈，虽然是一道字符串水题，可是拿到一个1A还是很开心的！
题意就是给一些keywords（子串）和Excuse（母串），然后输出包含keywords最多的Excuse，如果相等的话，按任意顺序全部输出即可。
解题时有几点需要注意：
1、一个keyword可能在Excuse里重复多次。
2、每个keyword的左右两端必须是行首或行尾或不是字母的字符。
3、Excuse里貌似是不区分大小写的，所以读入所有的Excuse后需要再memcpy一份，把那份用来全部转化为大写或小写，原来的那份作输出用。
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 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!
//#define LOCAL
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std; char key[][];
char excuse1[][], excuse2[][];
int sum[];//存放每个excuse中keywords出现的总次数 int match(char s1[], char s2[]); int main(void)
{
#ifdef LOCAL
freopen("409in.txt", "r", stdin);
#endif
int k, e, i, kase = , j;
while(scanf("%d %d", &k, &e) == )
{
++kase;
getchar();
for(i = ; i < k; ++i)
gets(key[i]);
for(i = ; i < e; ++i)
{
gets(excuse1[i]);
memcpy(excuse2[i], excuse1[i], sizeof(excuse1[i]));
for(j = ; j < strlen(excuse2[i]); ++j)
excuse2[i][j] = tolower(excuse2[i][j]);//将所有字母转化为小写
} memset(sum, , sizeof(sum));
int maxn = ;
for(i = ; i < e; ++i)
{
for(j = ; j < k; ++j)
{
sum[i] += match(excuse2[i], key[j]);
}
if(sum[i] > maxn)
maxn = sum[i];
} cout << "Excuse Set #" << kase << endl;
for(i = ; i < e; ++i)
{
if(sum[i] == maxn)
cout << excuse1[i] << endl;
}
cout << endl;
}
return ;
}
int match(char s1[], char s2[])
{
int i = , l1, l2, count = , j;
l1 = strlen(s1);
l2 = strlen(s2);
while(i + l2 < l1)//一个关键词可能匹配多次
{
int j = i, k = ;
if((i==||s1[i-]<'a'||s1[i-]>'z') &&
(i+l2==l1)||s1[i+l2]<'a'||s1[i+l2]>'z')//保证所匹配的单词能被分隔开
{
for(; k < l2; ++j, ++k)
{
if(s1[j] != s2[k])
break;
}
if(k == l2)
++count;
}
++i;
}
return count;
}
代码君