poj 1598 Excuses, Excuses!

//这题有两点需要注意的：1.有可能excuses中存在着大写字母，这就需要将其大写字母转换为小写字母再比较//2.keysword不能是excuses中的字串，而是一个单词，具体见下面的样例！ #include <iostream>#include <string>#include <cctype>#include <algorithm>using namespace std;struct Info{       int c;       string str;};bool mycmp(Info a, Info b){     return a.c > b.c;}int main(){    int i, j, k, e, pos, len, tc = 0, c;    string keysword[25];    Info excuses[25], tmp[25];    while (cin >> k >> e){          tc++;          //keysword和excuses的输入           for (i = 0; i < k; i++)               cin >> keysword[i];          cin.get();          for (i = 0; i < e; i++){               getline(cin, excuses[i].str);               excuses[i].c = 0;               tmp[i].str = excuses[i].str;               tmp[i].c = excuses[i].c;          }                    //讲excuses中的不是字母的转换为空格，而大写字母转换为小写，主要为后面的进行统一的比较           for (i = 0; i < e; i++){              len = excuses[i].str.length();              for (j = 0; j < len; j++){                  if (isalpha(excuses[i].str[j]))                      excuses[i].str[j] = tolower(excuses[i].str[j]);                  else                      excuses[i].str[j] = ' ';              }          }                    //找出keysword中excuses中的位置，如果是一个单词的情况下就符合情况，如果只是excuses中的一个字串的情况就不符合情况！           for (i = 0; i < e; i++){              for (j = 0; j < k; j++){                  pos = excuses[i].str.find(keysword[j]);                  if (pos >= 0){                      pos = pos + keysword[j].length();                      if (excuses[i].str[pos] == ' ')                          tmp[i].c++;                  }              }          }                    sort(tmp, tmp+e, mycmp);                    cout << "Excuse Set #" << tc << endl;          cout << tmp[0].str << endl;          i = 1;          while (tmp[i].c == tmp[0].c && i < e){                cout << tmp[i].str << endl;                i++;          }          cout << endl;    }        system("pause");}/*2 2aaabaaaa abaaaba aaaa abaaad*/