题解:SP28 HMRO - Help the Military Recruitment Office!
题意
已知
思路
按照题目模拟。使用两个 map<string,string> 分别为 m1 和 fa.
m1存储每个人的 PIESL 和旧的 MRO。fa存储旧的 MRO 和新的 MRO。
使用并查集进行优化。
代码
#include<bits/stdc++.h>
#define int long long
using namespace std;
int T, n, m, t;
map<string, string> m1, fa;
string findfa(string x) {
return fa[x] = (fa[x] == x ? x : findfa(fa[x]));
}
signed main() {
cin.tie(0), cout.tie(0);
cin >> T;
while (T--) {
m1.clear(), fa.clear();
cin >> n;
for (int i = 1; i <= n; i++) {
string PIESL, MRO;
cin >> PIESL >> MRO;
m1[PIESL] = MRO;
fa[MRO] = MRO;
}
cin >> m;
for (int i = 1; i <= m; i++) {
string Old, New;
cin >> Old >> New;
fa[findfa(Old)] = findfa(New);
}
cin >> t;
for (int i = 1; i <= t; i++) {
string q;
cin >> q;
cout << q << " " << findfa(m1[q]) << "\n";
}
}
return 0;
}
附上 AC 记录