这个题比较长，如果按照题目要求一步一步做的话。也不是很难。但是如果想提高代码精简度和可读性高效性确实不容易，想轻松AC还是用C++，C语言的输入输出。Java很容易超时。普通算法要10s左右。
Java代码 ：
import java.util.*;
public class Main1597 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
String s = scan.nextLine();
String[][] str = new String[105][1500];
int[] row = new int[t];
Arrays.fill(row, 0);
for(int i=0; i<t; i++) {
while(true) {
str[i][row[i]] = scan.nextLine();
if(str[i][row[i]].equals("**********")) {
break;
}
row[i] ++;
}
}
int n = scan.nextInt();
String ss = scan.nextLine();
for(int j=0; j<n; j++) {
String line = scan.nextLine();
String[][] ans = new String[t][1000];
int[] cnt = new int[1000];
Arrays.fill(cnt, 0);
if(line.contains(" AND ")) {
int end = 0;
while(line.charAt(end)!=' ')end ++;
String A = line.substring(0, end);
String B = line.substring(end+5, line.length());
for(int i=0; i<t; i++) {
int a = 0, b = 0;
for(int r=0; r<row[i]; r++) {
if(str[i][r].toLowerCase().contains(A)) {
a = 1;
}
if(str[i][r].toLowerCase().contains(B)) {
b = 1;
}
}
if(a==1 && b==1) {
for(int r=0; r<row[i]; r++) {
if(str[i][r].toLowerCase().contains(A) || str[i][r].toLowerCase().contains(B)) {
ans[i][cnt[i]++] = str[i][r];
}
}
}
}
}else if(line.contains(" OR ")) {
int end = 0;
while(line.charAt(end)!=' ')end ++;
String A = line.substring(0, end);
String B = line.substring(end+4, line.length());
for(int i=0; i<t; i++) {
for(int r=0; r<row[i]; r++) {
if(str[i][r].toLowerCase().contains(A) || str[i][r].toLowerCase().contains(B)) {
ans[i][cnt[i]++] = str[i][r];
}
}
}
}else if(line.contains("NOT ")) {
int end = 0;
while(line.charAt(end)!=' ')end ++;
//String A = line.substring(0, end);
String B = line.substring(4, line.length());
for(int i=0; i<t; i++) {
int con = 0;
for(int r=0; r<row[i]; r++) {
if(!str[i][r].toLowerCase().contains(B)) {
//ans[i][cnt[i]++] = str[i][r];
con ++;
}
}
if(con == row[i]) {
for(int r=0; r<row[i]; r++) {
ans[i][cnt[i]++] = str[i][r];
}
}
}
}else {
for(int i=0; i<t; i++) {
for(int r=0; r<row[i]; r++) {
if(str[i][r].toLowerCase().contains(line)) {
ans[i][cnt[i]++] = str[i][r];
}
}
}
}
int flag = 0, m = 0;
for(int k=0; k<t; k++) {
if(m > 0 && cnt[k] != 0)
System.out.println("----------");
for(int l=0; l<cnt[k]; l++) {
//System.out.println("----------");
System.out.println(ans[k][l]);
flag = 1;
m ++;
}
}
if(flag == 0)
System.out.println("Sorry, I found nothing.");
System.out.println("==========");
}
}
}
来自互联网的C++代码 ：
//{头文件模板
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
using namespace std;
//}头文件模板
#define rep(i,b) for(int i=0; i<(b); i++)
#define foreach(i,a) for(__typeof((a).begin()) i=a.begin(); i!=(a).end(); ++i)
#define FOR for (int j = limit[i]; j < limit[i + 1]; j++)
typedef bool Bit[1505];
int n, lines, m; // n文档数, lines行数, m请求数
int limit[105]; // limit[i]: 第i篇文档从第几行开始
string doc[1505]; // 存储内容
map<string, Bit> Index; // Index[单词]: B标记了哪些行出现过
// 用s来更新Index
void upDate(string s, int p) {
string word;
foreach(it, s) {
if (isalpha(*it)) *it = tolower(*it); // 变小写
else *it = ' '; // 非字母变空白
}
stringstream ss(s);
while (ss >> word) Index[word][p] = true;
}
int main() {
//{section: 文档数据输入
scanf("%d ", &n);
rep(i, n) {
limit[i] = lines;
while (getline(cin, doc[lines]), doc[lines] != "**********") {
upDate(doc[lines], lines);
lines++;
}
}
limit[n] = lines;//}
//{section: 对获取的请求输出对应的内容
string s; // s存储每次得到的请求
Bit mark; //{记录哪些行应该输出
//}mark的解释: 因为每篇文档要用10个'-'隔开,比较麻烦,最后统一处理
bool *A, *B;
scanf("%d ", &m);
for (int iii = 0; iii < m; iii++) {
getline(cin, s);
//{subsection: 计算出mark中介
if (s[0] == 'N') {
A = Index[s.substr(4)];
rep(i, n) {
bool logo = true;
FOR if (A[j]) { logo = false; break; }
FOR mark[j] = logo;
}
} else if (s.find("AND") != string::npos) {
int p = s.find(" AND ");
A = Index[s.substr(0, p)];
B = Index[s.substr(p + 5)];
memset(mark, 0, sizeof(mark));
bool hasA, hasB; // 在同一文章中, 两个词是否都出现
rep(i ,n) {
hasA = hasB = false; // 默认没出现
FOR if (A[j]) { hasA = true; break; }
FOR if (B[j]) { hasB = true; break; }
if (!(hasA&&hasB)) continue;
FOR mark[j] = (A[j] || B[j]);
}
} else if (s.find("OR") != string::npos) {
int p = s.find(" OR ");
A = Index[s.substr(0, p)];
B = Index[s.substr(p + 4)];
rep(i, lines) mark[i] = (A[i] || B[i]);
} else memcpy(mark, Index[s], sizeof(mark));//}
//{subsection: 输出mark
bool hasOut = false, needOut = false; // 记录上一轮是否有输出文档
rep(i, n) {
if (hasOut) needOut = true;
hasOut = false;
FOR if (mark[j]) {
if (needOut) {
cout << "----------
";
needOut = false;
}
cout << doc[j] << "
";
hasOut = true;
}
}
if (!(needOut||hasOut)) cout << "Sorry, I found nothing.
";
cout << "==========
";//}
}//}
return 0;
}