이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int mxN=2e3, mxC=50;
int n, m, ci, fi, vi;
ll dp[mxN+1][mxC*2+1], ans;
vector<array<int, 3>> v1, v2;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n;
for(int i=0; i<n; ++i) {
cin >> ci >> fi >> vi;
v1.push_back({fi, ci, vi});
}
sort(v1.begin(), v1.end(), greater<array<int, 3>>());
cin >> m;
for(int i=0; i<m; ++i) {
cin >> ci >> fi >> vi;
v2.push_back({fi, ci, vi});
}
sort(v2.begin(), v2.end(), greater<array<int, 3>>());
memset(dp, 0x96, sizeof(dp));
dp[0][50]=0;
for(int i=0; i<=n; ++i) {
for(int j=0; j<=m; ++j) {
for(int k=2*mxC; k>=0; --k) {
if(k<mxC&&i<n&&j&&v1[i][0]>=v2[j-1][0])
dp[j][k+v1[i][1]]=max(dp[j][k]-v1[i][2], dp[j][k+v1[i][1]]);
if(k>=mxC&&j<m) {
dp[j+1][k-v2[j][1]]=max(dp[j][k]+v2[j][2], dp[j+1][k-v2[j][1]]);
dp[j+1][k]=max(dp[j][k], dp[j+1][k]);
}
}
}
}
for(int i=mxC; i<=2*mxC; ++i)
ans=max(dp[m][i], ans);//, cout << i << " " << dp[m][i] << endl;
cout << ans;
}
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