第四题（共四题100分）：低频词过滤（40分）
题目描述：请编写程序，从包含大量单词的文本中删除出现次数最少的单词。如果有多个单词都出现最少的次数，则将这些单词都删除。
输入数据：程序读入已被命名为corpus.txt的一个大数据量的文本文件，该文件包含英文单词和中文单词，词与词之间以一个或多个whitespace分隔。（为便于调试，您可下载测试corpus.txt文件，实际运行时我们会使用不同内容的输入文件。）
输出数据：在标准输出上打印删除了corpus.txt中出现次数最少的单词之后的文本（词与词保持原来的顺序，仍以空格分隔）。
评分标准：程序输出结果必须正确，内存使用越少越好，程序的执行时间越快越好。
我的程序：
#pragma warning(disable: 4786)#include <map>#include <string>#include <vector>#include <cstdio>#include <fstream>#include <algorithm>using namespace std;struct Pair
{
Pair(){}
Pair()
Pair(const string& w, int c):word(w), count(c) {
} string word; int count;
};map< string, int > wordmap;vector< Pair > wordcount;vector< int > wordindex;int main(){ // load the file, count word, build index string word; fstream file("corpus.txt"); while(file >> word) { int id; map< string, int >::iterator itw = wordmap.find(word); if (itw == wordmap.end()) // new word { id = wordcount.size(); wordmap[word] = id; wordcount.push_back(Pair(word, 1)); // 奇怪,我当时写的是id吗？- -晕了，谢谢mammger 指出来 } else // found it { wordcount[itw->second].count++; id = itw->second; } wordindex.push_back(id); } file.close(); if (wordcount.empty()) return 0; // find the min count vector< Pair >::iterator itc = wordcount.begin(); int mincount = itc->count; for (++itc; itc != wordcount.end(); ++itc) { if (mincount > itc->count) mincount = itc->count; } // show all filtered words if (wordindex.empty()) return 0; // skip leading filtered words vector< int >::iterator w = wordindex.begin(); while(w != wordindex.end() && wordcount[*w].count == mincount) ++w; if (w == wordindex.end()) return 0; /* debug use for (vector< Pair >::iterator iw = wordcount.begin(); iw != wordcount.end(); ++iw) printf("%s %d\n", iw->word.c_str(), iw->count); printf("mincount = %d\n", mincount); /*/ // show first word, with no space after printf("%s", wordcount[*w].word.c_str()); for (++w; w != wordindex.end(); ++w) { if (wordcount[*w].count != mincount) { printf(" %s", wordcount[*w].word.c_str()); } } printf("\n"); //*/ return 0;}