思路：
使用堆即可。
实现：
1 #include <iostream> 2 #include <map> 3 #include <vector> 4 #include <cstring> 5 #include <queue> 6 #include <algorithm> 7 8 using namespace std; 9 10 typedef pair<int, int> pii; 11 12 int buf[6], need[6][10005]; 13 14 int main() 15 { 16 int T, n, m; 17 cin >> T; 18 map<string, int> mp; 19 mp["CV"] = 0; mp["DD"] = 1; mp["CL"] = 2; 20 mp["CA"] = 3; mp["BB"] = 4; mp["BC"] = 5; 21 while (T--) 22 { 23 memset(need, 0, sizeof need); 24 cin >> n >> m; 25 string s; 26 int l, f; 27 vector<pii> v[6]; 28 for (int i = 0; i < n; i++) 29 { 30 cin >> s >> l >> f; 31 v[mp[s]].push_back(pii(l, f)); 32 } 33 for (int i = 0; i < 6; i++) sort(v[i].begin(), v[i].end()); 34 for (int i = 0; i < m; i++) 35 { 36 for (int j = 0; j < 6; j++) cin >> buf[j]; 37 cin >> l; 38 for (int j = 0; j < 6; j++) 39 { 40 need[j][l] += buf[j]; 41 } 42 } 43 bool flg = true; 44 int ans = 0; 45 for (int i = 0; i < 6; i++) 46 { 47 priority_queue<int, vector<int>, greater<int>> q; 48 for (int j = 10000; j >= 1; j--) 49 { 50 if (need[i][j]) 51 { 52 while (!v[i].empty() && v[i].back().first >= j) 53 { 54 q.push(v[i].back().second); 55 v[i].pop_back(); 56 } 57 if (q.size() < need[i][j]) { flg = false; break; } 58 while (need[i][j]) { ans += q.top(); q.pop(); need[i][j]--; } 59 } 60 } 61 if (!flg) break; 62 } 63 if (!flg) cout << -1 << endl; 64 else cout << ans << endl; 65 } 66 return 0; 67 }