Submission #1222745

#	Time	Username	Problem	Language	Result	Execution time	Memory
1222745		sitingfake	Cloud Computing (CEOI18_clo)	C++20	0 / 100	23 ms	2120 KiB

clo

#include<bits/stdc++.h>
using namespace std;

// define

#define execute cerr << " Time: " << fixed << setprecision(6) << (1.0 * clock() / CLOCKS_PER_SEC) << "s\n";
#define ll long long
#define ld double
#define ii pair<int,int>
#define se second
#define fi first
#define iii pair<int,ii>
#define all(v) v.begin(),v.end()
#define bit(x,i) (((x)>>(i))&1LL)
#define flip(x,i) ((x)^(1LL<<(i)))
#define ms(d,x) memset(d,x,sizeof(d))
#define sitingfake 1
#define orz 1

//constant

const ll mod = 1e9 + 7;
const long long linf = 4557430888798830399;
const int inf = 1061109567;
const int maxarr = 1e6 + 5;
const int dx[] = {0,1,-1,0};
const int dy[] = {1,0,0,-1};

template<typename T> bool maximize(T &a, const T &b)
{
    if(a < b) {a = b; return 1;}
    return 0;
}

template<typename T> bool minimize(T &a, const T &b)
{
    if(a > b) {a = b; return 1;}
    return 0;
}

inline void Plus(ll & a ,ll b)
{
    b %= mod;
    a += b;
    if(a >= mod) a -= mod;
    if(a < 0) a += mod;
    return;
}

inline void Mul(ll & a, ll b)
{
    a %= mod; b %= mod;
    a *= b;
    a %= mod;
    return;
}

//code

const int maxn = 2e3 + 5;

struct Computer
{
    int c;
    int f;
    int val;

    bool operator < (Computer &Other)
    {
        if(f == Other.f) return val < Other.val;
        return f < Other.f;
    }
};

struct Customer
{
    int c;
    int f;
    int val;

    bool operator < (Customer &Other)
    {
        if(f == Other.f) return val > Other.val;
        return f < Other.f;
    }
};

Computer a[maxn];
Customer b[maxn];

ll dp[2][maxn][55];

int prec[maxn];

int PreV[maxn];

int n , m;

void solve(void)
{
    cin >> n;

    for(int i = 1; i <= n; i++)
    {
        cin >> a[i].c >> a[i].f >> a[i].val;
    }

    cin >> m;

    for(int i = 1; i <= m; i++)
    {
        cin >> b[i].c >> b[i].f >> b[i].val;
    }

    sort(a + 1 , a + n + 1);
    sort(b + 1 , b + m + 1);

    for(int i = 1; i <= n; i++)
    {
        prec[i] = a[i].c + prec[i-1];
        PreV[i] = a[i].val + PreV[i-1];
    }

    ms(dp , -0x3f);

    for(int i = 1; i <= n; i++)
    {
        dp[0][i][0] = 0;
    }

    for(int i = 1; i <= m; i++)
    {
        int cur = i & 1;
        int pre = !cur;
        ms(dp[cur] , -0x3f);

        for(int j = 1; j <= n; j++)
        {
            for(int k = 0; k <= a[j].c; k++)
            {
                // don't take missions i
                dp[cur][j][k] = dp[pre][j][k];
            }
            // don't take computer j

            maximize(dp[cur][j][0] , dp[pre][j-1][a[j-1].c]);

            // we take mission i

            if(a[j].f >= b[i].f)
            {
                for(int k = 1; k <= a[j].c; k++)
                {
                    int l = 1 , r = j - 1 , pos = -1;

                    if(k > b[i].c)
                    {
                        maximize(dp[cur][j][k] , dp[pre][j][k - b[i].c] + b[i].val);
                    }
                    else if(k == b[i].c)
                    {
                        maximize(dp[cur][j][k] , dp[pre][j][0] - a[j].val + b[i].val);
                    }
                    else
                    {
                        while(l <= r)
                        {
                            int mid = (r + l) >> 1;
                            if(prec[j-1] - prec[mid-1] >= b[i].c - k)
                            {
                                pos = mid;
                                l = mid + 1;
                            }
                            else r = mid - 1;
                        }

                        if(pos == -1 || a[pos].f < b[i].f) continue;

                        int val = b[i].c - k - (prec[j-1] - prec[pos]);

                        if(a[pos].c - val > 0)
                        {
                            maximize(dp[cur][j][k] , dp[pre][pos][a[pos].c - val] - PreV[j] + PreV[pos] + b[i].val);
                        }

                        maximize(dp[cur][j][k] ,  dp[pre][pos][0] - PreV[j] + PreV[pos-1] + b[i].val);
                    }
                }
            }

            for(int k = 2; k <= a[j].c; k++)
            {
                maximize(dp[cur][j][k] , dp[cur][j][k-1]);
            }

            for(int k= a[j].c + 1; k <= 50; k++) dp[cur][j][k] = -linf;
        }
    }

    ll ans = 0;

    for(int i = 1; i <= n; i++)
    {
        for(int k = 0; k <= a[i].c; k++)
        {
            maximize(ans , dp[m&1][i][k]);
        }
    }

    cout << ans;


}
/**
4
2 1800 10
4 2000 750
4 2200 700
20 2550 9999
3
1 1500 300
6 1900 1500
3 2400 4550
**/
signed main()
{
   ios_base::sync_with_stdio(0);
   cin.tie(0);
   cout.tie(0);

   #define task ""

   if(fopen(task".inp","r"))
   {
       freopen(task".inp","r",stdin);
       freopen(task".out","w",stdout);
   }

   int tc; tc = 1;

   while(tc--) solve();

   //execute;
}

Compilation message (stderr)

clo.cpp: In function 'int main()':
clo.cpp:235:15: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  235 |        freopen(task".inp","r",stdin);
      |        ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~
clo.cpp:236:15: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  236 |        freopen(task".out","w",stdout);
      |        ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~

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