难度中等33
每年,政府都会公布一万个最常见的婴儿名字和它们出现的频率,也就是同名婴儿的数量。有些名字有多种拼法,例如,John 和 Jon 本质上是相同的名字,但被当成了两个名字公布出来。给定两个列表,一个是名字及对应的频率,另一个是本质相同的名字对。设计一个算法打印出每个真实名字的实际频率。注意,如果 John 和 Jon 是相同的,并且 Jon 和 Johnny 相同,则 John 与 Johnny 也相同,即它们有传递和对称性。
在结果列表中,选择 字典序最小 的名字作为真实名字。
示例:
输入:names = ["John(15)","Jon(12)","Chris(13)","Kris(4)","Christopher(19)"], synonyms = ["(Jon,John)","(John,Johnny)","(Chris,Kris)","(Chris,Christopher)"] 输出:["John(27)","Chris(36)"]
提示:
names.length <= 100000
class Solution {
public:
int f[100001];
int rank[100001];
void Make_Set(int N)
{
for (int i = 0; i < N; i++)
{
f[i] = i;
rank[i] = 0;
}
}
int findF(int x) {
return f[x] == x ? x : f[x] = findF(f[x]);
}
void merge(int x, int y) {
int fx = findF(x), fy = findF(y);
if (fx == fy) return;
if (rank[fx] < rank[fy]) {
swap(fx, fy);
}
rank[fx] += rank[fy];
f[fy] = fx;
}
vector<string> trulyMostPopular(vector<string>& names, vector<string>& synonyms) {
int N = names.size();
unordered_map<string, int>room;
unordered_map<int, string>rroom;
vector<int>fre(N);
vector<string>res;
Make_Set(N);
if (names.size() == 0)
return res;
for (int i = 0; i < N;++i) {
int l = names[i].find('(');
names[i][l] = '*';
int r = names[i].find(')');
string name = names[i].substr(0, l);
int val = stoi(names[i].substr(l + 1, r - l + 1));;
room[name] = i;
rroom[i] = name;
fre[i] = val;
//cout << name << ' ' << fre[i] << endl;
}
for (auto s : synonyms) {
int l = s.find(',');
string a = s.substr(1, l - 1);
string b = s.substr(l + 1, s.size() - l - 2);
merge(room[a], room[b]);
//cout << a << ' ' << b << endl;
}
for (int i = 0; i < N; ++i) {
int fa = findF(i);
if (fa != i) {
if (rroom[i] < rroom[fa]) {
rroom[fa] = rroom[i];
}
fre[fa] += fre[i];
}
}
for (int i = 0; i < N; ++i) {
int fa = findF(i);
if (i == fa) {
string name = rroom[i];
int num = fre[i];
name += "(" + to_string(num) + ")";
//cout << name << endl;
res.push_back(name);
}
}
return res;
}
};