Submission #947160

#	Time	Username	Problem	Language	Result	Execution time	Memory
947160		DJeniUp	Cloud Computing (CEOI18_clo)	C++17	18 / 100	727 ms	2288 KiB

clo

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll>pairll;
typedef pair<ll,ull>pairull;
typedef pair<ll,pairll>pair3l;
typedef long double ld;
typedef pair<ld,ld>pairld;

typedef pair<string,ll>pairsl;

#define fr first
#define sc second
#define pb push_back
#define endl '\n'
#define N 100007
#define MOD 1000000007
#define INF 10000000000007
#define eps 0.00000000001
#define A 49
#define PI 3.14159265359

ll n,m,dp[2][2007][A+7];

struct D{
    ll c,f,v;
}d[2007],t[2007];

bool mcp(D d1, D d2){
    return d1.f<d2.f;
}

int main(){

    ios_base::sync_with_stdio(0);
    cout.tie(0);
    cin.tie(0);

    cin>>m;
    for(int i=1;i<=m;i++){
        cin>>t[i].c>>t[i].f>>t[i].v;
    }
    sort(t+1,t+1+m,mcp);
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>d[i].c>>d[i].f>>d[i].v;
    }
    sort(d+1,d+1+n,mcp);
    for(int i=0;i<2;i++){
        for(int j=1;j<=m+1;j++){
            for(int k=1;k<=A+1;k++){
                dp[i][j][k]=-INF;
            }
        }
    }
    d[n+1].f=d[n].f;
    for(int i1=n+1;i1>=1;i1--){
        swap(dp[0],dp[1]);
        /*for(int j=1;j<=m+1;j++){
            for(int k=1;k<=A+1;k++){
                dp[0][j][k]=-INF;
            }
        }*/
        ll i=0;
        for(int j=m;j>=1;j--){
            for(int k=max(d[i1].c,max(d[i1-1].c,t[j].c));k>=0;k--){
               // cout<<i<<" "<<j<<" "<<k<<endl;
                dp[i][j][k]=max(dp[i+1][j][k],max(dp[i][j+1][k],dp[i][j][k+1]));
                if(t[j].f>=d[i1].f && k+d[i1].c-t[j].c>=0 && k+d[i1].c-t[j].c<=A)dp[i][j][k]=max(dp[i][j][k],dp[i+1][j+1][k+d[i1].c-t[j].c]+d[i1].v-t[j].v);
                if(t[j].f>=d[i1-1].f && k-t[j].c>=0)dp[i][j][k]=max(dp[i][j+1][k-t[j].c]-t[j].v,dp[i][j][k]);
                if(k+d[i1].c<=A)dp[i][j][k]=max(dp[i][j][k],dp[i+1][j][k+d[i1].c]+d[i1].v);
                //if(k<=30)cout<<i<<" "<<j<<" "<<k<<" "<<dp[i][j][k]<<endl;
            }
        }
    }
    cout<<dp[0][1][0]<<endl;

    return 0;
}

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