제출 #345047

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
345047		ant101	Kitchen (BOI19_kitchen)	C++14	100 / 100	177 ms	206644 KiB

kitchen

#include <iostream>
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <fstream>
#include <cmath>
#include <vector>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <map>
#include <stack>
#include <queue>
#include <assert.h>
#include <limits>
#include <cstdio>
#include <complex>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef double db;
typedef string str;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
#define mp make_pair
#define f first
#define s second

typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<ld> vd;
typedef vector<str> vs;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<pd> vpd;

#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend()
#define rsz resize
#define ins insert
#define ft front()
#define bk back()
#define pf push_front
#define pb push_back
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353; // = (119<<23)+1
const int MX = 3e2+5;
const ll INF = 1e18;
const ld PI = 4*atan((ld)1);
const int dx[4] = {0,1,0,-1}, dy[4] = {1,0,-1,0};


namespace io {
    void setIn(string s) { freopen(s.c_str(),"r",stdin); }
    void setOut(string s) { freopen(s.c_str(),"w",stdout); }
    void setIO(string s = "") {
        ios_base::sync_with_stdio(0); cin.tie(0); // fast I/O
        // cin.exceptions(cin.failbit); // ex. throws exception when you try to read letter into int
        if (sz(s)) {
            setIn(s+".in");
            setOut(s+".out");
        } // for USACO
    }
}

using namespace io;

namespace input {
    template<class T> void re(complex<T>& x);
    template<class T1, class T2> void re(pair<T1,T2>& p);
    template<class T> void re(vector<T>& a);
    template<class T, size_t SZ> void re(array<T,SZ>& a);
    
    template<class T> void re(T& x) { cin >> x; }
    void re(double& x) { string t; re(t); x = stod(t); }
    void re(ld& x) { string t; re(t); x = stold(t); }
    template<class Arg, class... Args> void re(Arg& first, Args&... rest) {
        re(first); re(rest...);
    }
    
    template<class T> void re(complex<T>& x) { T a,b; re(a,b); x = cd(a,b); }
    template<class T1, class T2> void re(pair<T1,T2>& p) { re(p.f,p.s); }
    template<class T> void re(vector<T>& a) { F0R(i,sz(a)) re(a[i]); }
    template<class T, size_t SZ> void re(array<T,SZ>& a) { F0R(i,SZ) re(a[i]); }
}

namespace output {
    template<class T1, class T2> void pr(const pair<T1,T2>& x);
    template<class T, size_t SZ> void pr(const array<T,SZ>& x);
    template<class T> void pr(const vector<T>& x);
    template<class T> void pr(const set<T>& x);
    template<class T1, class T2> void pr(const map<T1,T2>& x);
    
    template<class T> void pr(const T& x) { cout << x; }
    template<class Arg, class... Args> void pr(const Arg& first, const Args&... rest) {
        pr(first); pr(rest...);
    }
    
    template<class T1, class T2> void pr(const pair<T1,T2>& x) {
        pr("{",x.f,", ",x.s,"}");
    }
    template<class T> void prContain(const T& x) {
        pr("{");
        bool fst = 1; for (const auto& a: x) pr(!fst?", ":"",a), fst = 0; // const needed for vector<bool>
        pr("}");
    }
    template<class T, size_t SZ> void pr(const array<T,SZ>& x) { prContain(x); }
    template<class T> void pr(const vector<T>& x) { prContain(x); }
    template<class T> void pr(const set<T>& x) { prContain(x); }
    template<class T1, class T2> void pr(const map<T1,T2>& x) { prContain(x); }
    
    void ps() { pr("\n"); }
    template<class Arg> void ps(const Arg& first) {
        pr(first); ps(); // no space at end of line
    }
    template<class Arg, class... Args> void ps(const Arg& first, const Args&... rest) {
        pr(first," "); ps(rest...); // print w/ spaces
    }
}

using namespace output;

using namespace input;

ll add(ll a, ll b) {
    a += b;
    if(a >= MOD) {
        a -= MOD;
    }
    return a;
}
ll sub(ll a, ll b) {
    a -= b;
    if(a < 0) {
        a += MOD;
    }
    return a;
}

ll mul(ll a, ll b) {
    return (a * b)%MOD;
}

void add_self(ll& a, ll b) {
    a = add(a, b);
}
void sub_self(ll& a, ll b) {
    a = sub(a, b);
}
void mul_self(ll& a, ll b) {
    a = mul(a, b);
}

ll N, M, K;
ll mealtimes[MX];
ll dp[MX][MX*MX];
ll chefs[MX];

int main() {
//    ps(sizeof(dp)+sizeof(mealtimes)+sizeof(chefs));
    setIO();
    re(N, M, K);
    ll tmh = 0;
    F0R (i, N) {
        re(mealtimes[i]);
        if (mealtimes[i] < K) { // if we cannot even have K chefs on a certain dish then its legit impossible lmao
            ps("Impossible");
            return 0;
        }
        tmh+=mealtimes[i];
    }
    F0R (i, M) {
        re(chefs[i]);
    }
    // so lets see are we doing forward or backwards dp
    // our state is f(current index of the chefs array, sum of all the hours of the chefs that we have chosen)
    // value will be the value of max(sum(min(chefs[j], N))) over all used chefs j.
    // the idea is that in order to have K chefs on each dish, we need sum(min(chefs[j], N)) >= N*K
    // reason is because each chef can contribute to a maximum of N total dishes or less. We want the total contribution to be N*K, so each dish out of N can have K chefs on it
    // that reminds me i forgot about an edgecase :rhulkmoment:
    // if we let the minimum sum of all hours such that above condition is satisfied S, then the answer is S-tmh
    // so the answer will be the minimum value of (total of all mealtimes)-sum of all the hours of the chefs that we have chosen
    // this variant of dp is looking like knapsack
    // we will do forwards knapsack, in which case we must 1 index???
    // no i guess not
    // what are our initial conditions
    // I think leaving everything to 0 will suffice
    // nvm it doesnt
    F0R (i, M+1) {
        F0R (j, N*300+1) {
            dp[i][j] = -1e13;
        }
    }
    dp[0][0] = 0; // initial condition is that we start out with 0 total contribution
    F0R (i, M) {
        F0R (j, N*300+1) {
            if (dp[i][j] == -1e13) { // if its impossible then we dont care
                continue;
            }
//            ps(i, j, dp[i][j]);
            dp[i+1][j+chefs[i]] = max(dp[i+1][j+chefs[i]], dp[i][j]+min(chefs[i], N)); // first we try to include it in the subset of chefs
            dp[i+1][j] = max(dp[i+1][j], dp[i][j]); // or we dont include it
        }
    }
    FOR (i, tmh, N*300+1) {
        if (dp[M][i] >= N*K) {
            ps(i-tmh);
            return 0;
        }
    }
    ps("Impossible");
    return 0;
}

컴파일 시 표준 에러 (stderr) 메시지

kitchen.cpp: In function 'void io::setIn(std::string)':
kitchen.cpp:67:35: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
   67 |     void setIn(string s) { freopen(s.c_str(),"r",stdin); }
      |                            ~~~~~~~^~~~~~~~~~~~~~~~~~~~~
kitchen.cpp: In function 'void io::setOut(std::string)':
kitchen.cpp:68:36: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
   68 |     void setOut(string s) { freopen(s.c_str(),"w",stdout); }
      |                             ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 9 / 9
• Subtask 2: 22 / 22
• Subtask 3: 20 / 20
• Subtask 4: 21 / 21
• Subtask 5: 28 / 28