编程之美-最短摘要

题目：给定一段英文描述，包含n个单词，用空格分开每个单词，再给定m个单词的关键字，请编程实现：找出这段英文描述中包含了所有关键字的最短子串，并输出。

分析：本题其实就是《编程之美》中的最短摘要生成。通过遍历所有可能的子串（包含了所有关键字的子串），找出这些子串中长度最短的一个，将其输出。这里可以加速的地方在于，找到一个包含了所有关键字的子串之后，如何用比较快的方式去找下一个包含了所有关键字的子串，如果使用回溯，复杂度太高。

下面的实现方法比较巧妙，借用一个set集合not_find_set存放未找到的关键字，用一个map容器存放已经找到的关键字，并将找到的关键字次数进行计数。假设当前的遍历范围是[begin,end]，并且在这个范围所有的关键字被找到，将begin指针依次向后挪，直到某个关键字x的出现次数为0，这样当前的最短字符串即可得到。接着，将x加入not_find_set，扫描指针end向后挪，找到下一个包含了所有关键字的最短字符串。

void FindMinLenAbstract()
{
	int n;	//number of document words
	int m;	//number of keywords

	while (cin>>n>>m) {

		// input
		vector<string> seq;
		while (n--)
			*back_inserter(seq) = *istream_iterator<string>(cin);

		set<string> kwords;
		while (m--)
			*inserter(kwords, kwords.end()) = *istream_iterator<string>(cin);

		// find shortest abstract
		typedef vector<string>::iterator Vsit;

		//q is current scan range, and r is min abstract range
		pair<Vsit, Vsit> q(seq.begin(), seq.begin()), r(seq.begin(),seq.end());

		//record words that not being found between q.first and q.second
		set<string> notfound = kwords;

		//record words with an associate appearance count
		//that being found between q.first and q.second
		map<string, int> found;

		for(;;) {

			//still have keyword not being found
			if (!notfound.empty()) {

				//all conditions have being considered
				if (q.second == seq.end())
					break;

				set<string>::iterator it = notfound.find(*q.second);

				//current word is an not-found word
				if (it != notfound.end()) {
					++found[*it];
					notfound.erase(it);
				}
				else {
					map<string, int>::iterator it2 = found.find(*q.second);
					if (it2 != found.end())
						++(it2->second);
				}

				//next keyword in sequence
				++(q.second);
			}

			//all keywords have being found
			else {

				// find an min range from q.first to q.second that
				// include all keywords.
				map<string, int>::iterator it = found.find(*q.first);
				if (it != found.end() && !--(it->second)) {
					size_t rlen = distance(r.first, r.second);
					size_t qlen = distance(q.first, q.second);
					if (!rlen || rlen > qlen) r = q;

					notfound.insert(it->first);
					found.erase(it);
				}

				++(q.first);
			}
		}

		// output
		if (r.second == seq.end() && r.first == seq.begin())
			cout<<"No abstract available.";
		else {
			if (r.first != seq.begin())
				cout<<"... ";
			for (bool first = true; r.first != r.second; ++r.first, first = false) {
				if (!first)
					cout<<' ';
				cout<<*r.first;
			}
			if (r.second != seq.end())
				cout<<" ...";
		}
		cout<<'\n';
	}
}