Description
利用串操作实现基于给定英文段落（100~200000）构造英文词典（英文单词按字典序排序）
Input
一篇文章（包含若干个单词，标点符号全部是标准的英文字符）。
Output
输出文章里面所有单词（按照字典序，每个单词一行）。
Sample Input
Hello world!
Sample Output
hello
world
HINT
考察知识点：串和大量单词数据的存储排序问题, 时间复杂度O(nlogn)或O(n^2)，空间复杂度O(n)
Append Code
析：这个题就把一些符号都处理了，然后再把单词加入set中，然后输出就好，注意这个题输出全是小写，第一次没注意。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <list>
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
set<string> sets;
set<string> :: iterator it;
char s[200005];
int main(){
while(gets(s) != NULL){
int x = 0, i;
for(i = 0; s[i]; ++i){
if(!isalpha(s[i])){
s[i] = 0;
if(i != x) sets.insert(s+x);
x = i+1;
}
else s[i] = towlower(s[i]);
}
if(i != x) sets.insert(s+x);
}
for(it = sets.begin(); it != sets.end(); ++it)
printf("%s
", it->c_str());
return 0;
}