ZOJ Problem Set - 1315 Excuses, Excuses!（字符串处理的好题）

Excuses, Excuses!

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB

---

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 (1 <= K <= 20) defines the number of keywords to be used in the search. The second number (1 <= E <= 20) 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 (1 <= L <= 20) 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 occurrence 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!

---

Source: South Central USA 1996

分析：

题意：

先给定若干关键字（都是小写字母，长度不超过20个字符），再给若干待匹配的字符串（包含各种字符，包括空格等，长度不超过70个字符），要求输出待匹配字符串中包含关键字最多的串（不区分大小写字母。可能有多个符合条件的字符串，要求全部输出，顺序随意）。关键字数量和待匹配字符串的数量都不超过20个。有多组测试数据。

考虑到字符数和长度都很小，可以直接暴力匹配。用结构体存储相关信息。需要耐心，比较锻炼编程能力。

ac代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
char c[75];
vector<string> v;
string s;
struct Info
{
    string s;//存储待匹配的字符串
    int n;//存储关键词匹配数
}info;
vector<Info> vv;
bool cmp(const Info &a,const Info &b)//自定义结构体比较函数，由大到小排序
{
    return a.n>b.n;
}
int main()
{
    int n,m,i,j,k;
    int p;
    string t;
    int num=0;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        //if(c) printf("\n");
        num++;
        v.clear();
        vv.clear();
        for(i=0;i<n;i++)
        {
            cin>>s;
            v.push_back(s);
        }
        cin.getline(c,75);//读掉换行，很重要！
        for(i=0;i<m;i++)
        {
            cin.getline(c,75);
            s=c;
            for(j=0;j<s.size();j++)//先做大写字母转小写字母操作，预处理。
            {
                if(s[j]>=65&&s[j]<=90)
                s[j]+=32;//string可以直接这样转换，char则必须s[j]=(char)(s[j]+32)
            }
            info.n=0;
            for(j=0;j<v.size();j++)//字符串匹配使用暴力匹配就可以
            {
                for(k=0;k<s.size()-v[j].size();k++)
                {
                    t="";
                    for(p=k;p<k+v[j].size();p++)
                    t+=s[p];
                    if(k==0&&v[j]==t&&(s[p]<'a'||s[p]>'z'))//如果子串在一行的开头，子串后一位不可以是字母
                    info.n++;
                    else if(k>0&&v[j]==t&&(s[p]<'a'||s[p]>'z')&&(s[k-1]<'a'||s[k-1]>'z'))
                    //如果子串不在一行的开头，子串的前一位和后一位都不能是字母。
                    info.n++;
                }
            }
            info.s=c;//字符串保持不变，不能用s
            vv.push_back(info);
        }
        sort(vv.begin(),vv.end(),cmp);
        printf("Excuse Set #%d\n",num);
        for(i=0;i<vv.size();i++)
        {
            if(i!=0&&vv[i].n<vv[i-1].n)
            break;
            else cout<<vv[i].s<<endl;
        }
        cout<<endl;
    }
    return 0;
}