这个题就是考察STL容器的综合运用,由于文章比较长并且请求很多,直接暴力扫肯定会超时,所以需要用map作个映射。
我的思路:1.用一个map<int,map<string,set<int>>>,第一个int代表第几篇文章,第二个map把单词及其行数对应起来,因为一个单词可能在同一行中出现两次,所以用了 一 个 set来存单词的行位置.
因为要输出相应的文本,所以用一个web数组存下所有的输入,行数用line计算。
lines存储每篇文章的起始行和终止行。
2.由于要查询的关键词都是由字母组成的,并且忽略大小写,所以可以把输入的字符串处理一下,把字母都变成小写,非字母变成空格,再用stringstream读取。
#include<iostream>
#include<string>
#include<cstring>
#include<vector>
#include<map>
#include<sstream>
#include<cstdio>
#include<set>
using namespace std;
map<int,map<string,set<int> > >Search; //int表示第几篇文章,第二个map中的string代表文章中的单词,最里面的int代表行数
vector<string>web; //存放所有文本
vector<pair<int,int> >lines;
string trans(string &s)
{
int len = s.length();
for(int i = 0; i < len; i++)
if(!isalpha(s[i])) s[i] = ' ';
else s[i] = tolower(s[i]);
return s;
}
int main()
{
int i,j,k,line = 1;
int n;
cin >> n;
pair<int,int>ha;
getchar();
string s1,s2;
for(i = 1; i <= n; i++){
ha.first = line-1;
while(1){
getline(cin,s1);
web.push_back(s1);
++line;
if(s1[0] == '*') {ha.second = line - 3; lines.push_back(ha); break;}
s2 = trans(s1);
stringstream ss(s2);
while(ss >> s1) {
if(!Search[i].count(s1)) Search[i][s1] = set<int>();
Search[i][s1].insert(line-2);
}
}
}
int m;
cin >> m;
getchar();
for(i = 1; i <= m; i++){
string s;
getline(cin,s);
int no = 1,first = 1;
int k1 = s.find(" AND "),k2 = s.find(" OR "),k3 = s.find("NOT ");
if(k1 != -1 || k2 != -1){
s1 = s.substr(0,(k1 > 0 ? k1 : k2));
s2 = s.substr(k1 > 0 ? k1+5 : k2+4);
set<int>cnt;
for(j = 1; j <= n; j++){
cnt.clear();
int x = 0, y = 0;
if(Search[j].count(s1)){
for(set<int>::iterator it = Search[j][s1].begin(); it != Search[j][s1].end(); ++it)
cnt.insert(*it);
x = 1;
}
if(Search[j].count(s2)){
for(set<int>::iterator it = Search[j][s2].begin(); it != Search[j][s2].end(); ++it)
cnt.insert(*it);
y = 1;
}
if((k1 != -1 && x && y) || (k2 != -1 && (x || y))){
if(!first) cout << "----------\n";first = no = 0;
for(set<int>::iterator it = cnt.begin(); it != cnt.end(); ++it)
cout << web[*it] << endl;
}
}
}
else if(k3 != -1){
s1 = s.substr(4);
for(j = 1; j <= n; j++)
if(!Search[j].count(s1)){
if(!first) cout << "----------\n";
for(k = lines[j-1].first; k <= lines[j-1].second; k++)
cout << web[k] << endl;
first = no = 0;
}
}
else{
for(j = 1; j <= n; j++){
if(Search[j].count(s)){
if(!first) cout << "----------\n";
no = first = 0;
for(set<int>::iterator it = Search[j][s].begin(); it != Search[j][s].end(); ++it)
cout << web[*it] << endl;
}
}
}
if(no)cout << "Sorry, I found nothing.\n";
cout << "==========\n";
}
return 0;
}