結果

問題 No.453 製薬会社
ユーザー chaemon
提出日時 2016-12-04 00:07:44
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 4,078 bytes
コンパイル時間 983 ms
コンパイル使用メモリ 100,468 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-11-27 14:41:40
合計ジャッジ時間 1,506 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 9
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

// #includes {{{
#include <algorithm>
#include <numeric>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <list>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>
#include <cmath>
using namespace std;
// }}}
// pre-written code {{{
#define REP(i,n) for(int i=0;i<(int)(n);++i)
#define RREP(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define FOR(i,c) for(__typeof((c).begin()) i=(c).begin();i!=(c).end();++i)
#define LET(x,a) __typeof(a) x(a)
//#define IFOR(i,it,c) for(__typeof((c).begin())it=(c).begin();it!=(c).end();++it,++i)
#define ALL(c) (c).begin(), (c).end()
#define MP make_pair
#define EXIST(e,s) ((s).find(e)!=(s).end())
#define RESET(a) memset((a),0,sizeof(a))
#define SET(a) memset((a),-1,sizeof(a))
#define PB push_back
#define DEC(it,command) __typeof(command) it=command
const int INF=0x3f3f3f3f;
typedef long long Int;
typedef unsigned long long uInt;
#ifdef __MINGW32__
typedef double rn;
#else
typedef long double rn;
#endif
typedef pair<int,int> pii;
/*
#ifdef MYDEBUG
#include"debug.h"
#include"print.h"
#endif
*/
// }}}
typedef vector<double> arr;
typedef vector<arr> matrix;
const double EPS = 1e-8;
enum { OPTIMAL, UNBOUNDED, NOSOLUTION, UNKNOWN };
struct two_stage_simplex {
    int N, M, st;
    matrix a;
    vector<int> s;
    two_stage_simplex(const matrix &A, const arr &b, const arr &c)
        : N(A.size()), M(A[0].size()), a(N+2, arr(M+N+1)), s(N+2), st(UNKNOWN) {
            for (int j = 0; j < M; ++j) a[N+1][j] = c[j]; // make simplex table
            for (int i = 0; i < N; ++i)
                for (int j = 0; j < M; ++j) a[i+1][j] = A[i][j];
            for (int i = 0; i < N; ++i) a[i+1][M+N] = b[i]; // add helper table
            for (int i = 0; i < N; ++i) a[ 0 ][i+M] = 1;
            for (int i = 0; i < N; ++i) a[i+1][i+M] = 1;
            for (int i = 0; i < N; ++i) s[i+1]      = i+M;
            for (int i = 1; i <= N; ++i)
                for (int j = 0; j <= N+M; ++j) a[0][j] += a[i][j];
            st = solve();
        }
    int status() const { return st; }
    double solution() const { return -a[0][M]; }
    double solution(arr &x) const {
        x.resize(M, 0);
        for (int i = 0; i < N; ++i)
            x[s[i+1]] = a[i+1].back();
        return -a[0][M];
    }
    int solve() {
        M += N; N += 1;
        solve_sub(); // solve stage one
        if (solution() > EPS) return NOSOLUTION;
        N -= 1; M -= N;
        swap(a[0], a.back()); a.pop_back(); // modify table
        for (int i = 0; i <= N; ++i) {
            swap(a[i][M], a[i].back());
            a[i].resize(M+1);
        }
        return solve_sub(); // solve stage two
    }
    int solve_sub() {
        int p, q;
        while (1) {
            //print();
            for (q = 0; q <= M && a[0][q] >= -EPS; ++q);
            for (p = 0; p <= N && a[p][q] <=  EPS; ++p);
            if (q >= M || p > N) break;
            for (int i = p+1; i <= N; ++i) // bland's care for cyclation
                if (a[i][q] > EPS)
                    if (a[i][M]/a[i][q] < a[p][M]/a[p][q] ||
                            (a[i][M]/a[i][q] == a[p][M]/a[p][q] && s[i] < s[q])) p = i;
            pivot(p, q);
        }
        if (q >= M) return OPTIMAL;
        else        return UNBOUNDED;
    }
    void pivot(int p, int q) {
        for (int j = 0; j <= N; ++j)
            for (int k = M; k >= 0; --k)
                if (j != p && k != q)
                    a[j][k] -= a[p][k]*a[j][q]/a[p][q];
        for (int j = 0; j <= N; ++j)
            if (j != p) a[j][q] = 0;
        for (int k = 0; k <= M; ++k)
            if (k != q) a[p][k] = a[p][k]/a[p][q];
        a[p][q] = 1.0;
        s[p] = q;
    }
};
/*
int main() {
    for (int n, m; cin >> n >> m; ) {
        arr c(n+m), b(m);
        for (int i = 0; i < n; ++i)
            cin >> c[i], c[i] *= -1;
        matrix A(m, arr(n+m));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j)
                cin >> A[i][j];
            A[i][n+i] = 1;
            cin >> b[i];
        }
        two_stage_simplex tss(A, b, c);
        double ans = -tss.solution() * m;
        printf("Nasa can spend %.0f taka.\n", ans + 0.5 - EPS);
    }
}
*/
int main(){
    int C,D;
    cin>>C>>D;
    matrix A(2,arr(4,0.0));
    A[0][0] = 3.0/4.0;
    A[0][1] = 2.0/7.0;
    A[1][0] = 1.0/4.0;
    A[1][1] = 5.0/7.0;
    A[0][3] = 1.0;
    A[1][4] = 1.0;
    arr b(2);
    b[0] = C;
    b[1] = D;
    arr c(2);
    c[0] = -1000.0;
    c[1] = -2000.0;
    two_stage_simplex tss(A,b,c);
    double ans = -tss.solution();
    printf("%.10lf\n",ans);
    return 0;
}
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