超级大模拟
直接用map吧string对应到编号上来,然后在开个数组把每个编号对应到每个可以互相转化区块上来,预处理出区块的最小值,使用时直接取最小是即可
代码
#include <cstdio>
#include <iostream>
#include <map>
#define ll long long
#define min(a,b) ((a<b)?a:b)
using namespace std;
inline ll read(){
ll x = 0; int zf = 1; char ch = ' ';
while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;
}
map<string, int> mp;
string s[100005];
int a[100005];
int dy[100005];
int val[100005];
int main(){
int n = read(), k = read(), m = read();
for (int i = 1; i <= n; ++i)
cin >> s[i];
for (int i = 1; i <= n; ++i)
a[i] = read(), mp[s[i]] = i;
int q;
for (int i = 1; i <= k; ++i)
val[i] = 2147483647;
for (int i = 1; i <= k; ++i){
q = read();
for (int j = 0; j < q; ++j){
int t = read();
dy[t] = i, val[i] = min(val[i], a[t]);
}
}
string c; ll ans = 0;
for (int i = 1; i <= m; ++i){
cin >> c;
ans += val[dy[mp[c]]];
}
printf("%lld", ans);
return 0;
}