제출 #290182

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
290182		groeneprof	Cloud Computing (CEOI18_clo)	C++14	0 / 100	99 ms	116984 KiB

clo

#include <bits/stdc++.h>
#define int long long
#define P(a) do{if(debug) cout<<a<<endl;} while(false)
#define H(a) P(#a<<": "<<a)
#define FR(i,a,b) for(int i = (a); i<(b); i++)
#define F(i,n) FR(i,0,n)
#define f first.first
#define c first.second
#define v second
#define ppiii pair<pair<int,int>, int>
const bool debug = 0;
const int inf = 1e9;
using namespace std;


int n,m;
signed main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin>>n;
	vector<ppiii> comp(n);
	F(i,n){
		cin>>comp[i].c>>comp[i].f>>comp[i].v;
	}
	cin>>m;
	vector<ppiii> task(m);
	F(i,m){
		cin>>task[i].c>>task[i].f>>task[i].v;
	}

	sort(comp.begin(), comp.end());
	sort(task.begin(), task.end());

	int dp[n+2][m+2][105];

	F(i,n+2) F(j,m+2) F(k,105) dp[i][j][k] = -inf;

	F(i,51) F(j,m+1){
		dp[n][j][i] = 0;
	}
	F(i,51) F(j,n+1){
		dp[j][m][i] =0;
	}

	for(int i = n-1; i>=0; i--){//the computer we are deciding to take or not
		for(int j = m-1; j>=0; j--){ //the task we are deciding to take or not
			for(int k = 100; k>=0; k--){ //the amount of cores still needed from the last task / - the amount of cores still left + 50
				if(k>=50){ //there are still cores needed for the last task
					dp[i][j][k] = max(dp[i+1][j][k-comp[i].c]-comp[i].v, dp[i+1][j][k]); //take or don't take the computer
				}
				else if(i>0){ //there are cores left from the last computer (i-1)
					
					if(comp[i-1].f < task[j].f){ 
						P("a");
						dp[i][j][k] = dp[i][j][50]; //the clockspeed from the last computer is not good enough
					}
					else{
						P("b");
						dp[i][j][k] = max(dp[i][j+1][k],dp[i][j+1][k+task[j].c]+task[j].v);
					}
					/*if(i==3 && j == 2 && k == 50-4) {
						H(dp[i][j+1][k]);
						H(dp[i][j+1][k+task[j].c]);
						H(task[j].v);
						H(dp[i][j][50]);
						H(dp[i][j][k]);
					}*/
				}
			}
		}
	}
	cout<<dp[0][0][50]<<endl;
	H(dp[2][2][50]);
	H(dp[3][2][50]);
	H(dp[3][2][46]);
	H(comp[2].v);
	/*while(true){
		int i,j,k;
		cin>>i>>j>>k;
		cout<<dp[i][j][k+50]<<endl;
	}*/
	return 0;
}

• Subtask 1: 0 / 18
• Subtask 2: 0 / 18
• Subtask 3: 0 / 18
• Subtask 4: 0 / 18
• Subtask 5: 0 / 18
• Subtask 6: 0 / 10