PAT (Advanced Level) Practice_1034 Head of a Gang (30 分)_字符串型的map构建

• 题目地址
• 题目分析：这道题目，考察的内容其实较为繁琐，但是理清思路后，也不见得繁琐了，注释见代码中。
• 我的代码：

  #include<iostream>
  #include<string>
  #include<map>
  using namespace std;
  
  int n, k, aa[2001][2001] = { 0 }, weight[2001] = { 0 };//极限情况下2000个不同名配对
  int id = 1, peo = 0, len = 0, headlen = 0, headid;
  bool bb[2001] = { false };
  
  map<string, int>stoint;//利用map来构建string和int的一一对应关系
  map<int, string>itos;//利用map来构建int和string的一一对应关系
  map<string, int>ans;
  
  void dfs(int ent) {
  	bb[ent] = true;
  	peo++, len += weight[ent];//将每个节点的所有权重相加 == 2倍的连通分量的所有权重和
  	if (weight[ent] > headlen)
  		headlen = weight[ent], headid = ent;
  	for (int i = 1; i < id; i++)
  		if (bb[i] == false && aa[ent][i])
  			dfs(i);
  }
  
  int main()
  {
  	string s1, s2;
  	scanf("%d %d", &n, &k);
  	for (int i = 0, t; i < n; i++)//建立一一对应关系
  	{
  		cin >> s1 >> s2 >> t;
  		if (stoint[s1] == 0) stoint[s1] = id, itos[id++] = s1;
  		if (stoint[s2] == 0) stoint[s2] = id, itos[id++] = s2;
  		aa[stoint[s1]][stoint[s2]] += t, aa[stoint[s2]][stoint[s1]] += t;
  	}
  	for (int i = 1; i < id; i++)//计算所有节点权重
  		for (int j = 1; j < id; j++)
  			if (j != i && aa[i][j])
  				weight[i] += aa[i][j];
  	int cnt = 0;
  	for (int i = 1; i < id; i++)
  		if (bb[i] == false)
  		{
  			dfs(i);
  			if (peo > 2 && len > 2 * k)//这里的len是每个节点的所有权重和，故k需*2
  			{
  				cnt++;
  				ans[itos[headid]] = peo;
  			}
  			peo = 0, len = 0, headlen = 0;
  		}
  
  	printf("%d\n", cnt);
  	for (auto it = ans.begin(); it != ans.end(); it++)
  		cout << (*it).first << ' ' << (*it).second << endl;
  
  	return 0;
  }