Mahmoud and Ehab and the message
Hazrat Ali
Mahmoud knows that the i-th word can be sent with cost ai. For each word in his message, Mahmoud can either replace it with another word of the same meaning or leave it as it is. Can you help Mahmoud determine the minimum cost of sending the message?
The cost of sending the message is the sum of the costs of sending every word in it.
The first line of input contains integers n, k and m (1 ≤ k ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of words in their language, the number of groups of words, and the number of words in Mahmoud's message respectively.
The second line contains n strings consisting of lowercase English letters of length not exceeding 20 which represent the words. It's guaranteed that the words are distinct.
The third line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) where ai is the cost of sending the i-th word.
The next k lines describe the groups of words of same meaning. The next k lines each start with an integer x (1 ≤ x ≤ n) which means that there are x words in this group, followed by x integers which represent the indices of words in this group. It's guaranteed that each word appears in exactly one group.
The next line contains m space-separated words which represent Mahmoud's message. Each of these words appears in the list of language's words.
The only line should contain the minimum cost to send the message after replacing some words (maybe none) with some words of the same meaning.
5 4 4
i loser am the second
100 1 1 5 10
1 1
1 3
2 2 5
1 4
i am the second
107
5 4 4
i loser am the second
100 20 1 5 10
1 1
1 3
2 2 5
1 4
i am the second
116
Solution
#include <bits/stdc++.h>
#define INF (int) 1e9
using namespace std;
int main() {
int n, k, m;
cin >> n >> k >> m;
map<string, int> wid;
for (int i = 0; i < n; i++) {
string w;
cin >> w;
wid[w] = i;
}
vector<int> c(n);
for (int i = 0; i < n; i++) {
cin >> c[i];
}
while (k--) {
int x;
cin >> x;
vector<int> g(x);
for (int i = 0; i < x; i++) {
cin >> g[i];
g[i]--;
}
int mnc = INF;
for (int i = 0; i < x; i++) {
mnc = min(mnc, c[g[i]]);
}
for (int i = 0; i < x; i++) {
c[g[i]] = mnc;
}
}
long long ans = 0;
vector<string> w(m);
for (int i = 0; i < m; i++) {
cin >> w[i];
ans += c[wid[w[i]]];
}
cout << ans << endl;
return 0;
}