| 25 | } |
| 26 | |
| 27 | ll solve(vector<box>& v) |
| 28 | { |
| 29 | int n = v.size(); |
| 30 | sort(v.begin(), v.end(), comp); |
| 31 | for(int j = 0; j <= 10000; j++) |
| 32 | { |
| 33 | if(v[0].si >= j) |
| 34 | dp[0][j] = v[0].val; |
| 35 | else dp[0][j] = 0; |
| 36 | } |
| 37 | |
| 38 | for(int i = 1; i < n; i++) |
| 39 | { |
| 40 | for(int j = 0; j <= 10000; j++) |
| 41 | { |
| 42 | ll op1 = dp[i-1][j]; |
| 43 | ll op2 = (v[i].si >= j) ? (v[i].val + ((j + v[i].wi > 10000)?0:dp[i-1][j+v[i].wi])) : 0; |
| 44 | dp[i][j] = max(op1, op2); |
| 45 | } |
| 46 | } |
| 47 | return dp[n-1][0]; |
| 48 | } |
| 49 | |
| 50 | int main() { |
| 51 | int n; |