Codeforces1458B Glass Half Spilled (DP)

题意

有(n)个杯子，第(i)个杯子的容量为(a_i)，初始时有(b_i)单位的水。

可以将水在杯子之间转移，转移时会有一半的损耗。

若允许随意转移，但最终只拿(k)杯水，问(k)个杯子里的水体积和最大能为多少。

对于(k = 1, 2, ..., n)依次输出答案。

解法

(O(n^4))动态规划。

假设取(k)杯水，那么最优的做法会是将剩下的水都往这(k)杯里倒，直到倒满。所以就是要最大化

假设(dp_{k, c})表示取(k)杯，杯子总容量为(c)，不经过转移的最大水量。那么加上转移操作的最大水量就是(min(c, dp_{c, k} + frac{sum - dp_{c, k}}{2}))。

而(dp_{k, c})就类似01背包问题的dp算一下，最后再枚举容量求最终答案即可。

AC代码

#include <bits/stdc++.h>
using namespace std;
 
using ll = int64_t;
using ull = uint64_t;
using VI = vector<int>;
using VL = vector<ll>;
using VVI = vector<vector<int>>;
using VVL = vector<vector<ll>>;
using PII = pair<int,int>;
using PLL = pair<ll, ll>;
 
#define REP(i, _, __) for (int i = (_); i < (__); ++i)
#define PER(i, _, __) for (int i = (_-1); i >= (__); --i)
#define FOR(i, _, __) for (int i = (_); i <= (__); ++i)
#define ROF(i, _, __) for (int i = (_); i >= (__); --i)
#define FE(v, V) for (const auto& v: V)

#define EB emplace_back
#define PB push_back
#define MP make_pair
#define FI first
#define SE second
#define SZ(x) (int((x).size()))
#define ALL(x) (x).begin(),(x).end()
#define LLA(x) (x).rbegin(),(x).rend()

const double PI = acos(-1.0);
   
namespace Backlight {
    const int __BUFFER_SIZE__ = 1 << 20;
    bool NEOF = 1;
    int __top;
    char __buf[__BUFFER_SIZE__], *__p1 = __buf, *__p2 = __buf, __stk[996];

    template<typename T>
    T MIN(T a, T b) { return min(a, b); }

    template<typename First, typename... Rest>
    First MIN(First f, Rest... r) { return min(f, MIN(r...)); }

    template<typename T>
    T MAX(T a, T b) { return max(a, b); }

    template<typename First, typename... Rest>
    First MAX(First f, Rest... r) { return max(f, MAX(r...)); }

    template<typename T>
    void updMin(T& a, T b) { if (a > b) a = b; }

    template<typename T>
    void updMax(T& a, T b) { if (a < b) a = b; }

    inline char nc() {
        return __p1 == __p2 && NEOF && (__p2 = (__p1 = __buf) + fread(__buf, 1, __BUFFER_SIZE__, stdin), __p1 == __p2) ? (NEOF = 0, EOF) : *__p1++;
    }
   
    template<typename T>
    inline bool read(T &x) {
        char c = nc();
        bool f = 0; x = 0;
        while (!isdigit(c)) c == '-' && (f = 1), c = nc();
        while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = nc();
        if (f) x = -x;
        return NEOF;
    }

    inline bool need(char c) { return (c != '
') && (c != ' '); }

    inline bool read(char& a) {
        while ((a = nc()) && need(a) && NEOF) ;
        return NEOF;
    }

    inline bool read(char *a) {
        while ((*a = nc()) && need(*a) && NEOF) ++a; 
        *a = '';
        return NEOF;
    }

    inline bool read(double &x) {
        bool f = 0; char c = nc(); x = 0;
        while (!isdigit(c))  { f |= (c == '-'); c = nc(); }
        while (isdigit(c)) { x = x * 10.0 + (c ^ 48); c = nc(); }
        if (c == '.') {
            double temp = 1; c = nc();
            while (isdigit(c)) { temp = temp / 10.0; x = x + temp * (c ^ 48); c = nc(); }
        }
        if (f) x = -x;
        return NEOF;
    }

    template<typename T, typename... T2>
    inline bool read(T &x, T2 &... rest) {
        read(x);
        return read(rest...);
    }

    template<typename T>
    inline void print(T x) {
        if (x < 0) putchar('-'), x = -x;
        if (x == 0) { putchar('0'); return; }
        __top = 0;
        while(x) {
            __stk[++__top] = x % 10 + '0';
            x /= 10;
        }
        while(__top) {
            putchar(__stk[__top]);
            --__top;
        }
    }

    template<typename First, typename... Rest>
    inline void print(First f, Rest... r) {
        print(f); putchar(' ');
        print(r...);
    }

    template<typename T>
    inline void println(T x) {
        print(x); 
        putchar('
');
    }

    template<typename First, typename... Rest>
    inline void println(First f, Rest... r) {
        print(f); putchar(' ');
        println(r...);
    }

    template<typename T>
    inline void _dbg(const char *format, T value) { cerr << format << '=' << value << endl; }
   
    template<typename First, typename... Rest>
    inline void _dbg(const char *format, First f, Rest... r) {
        while(*format != ',') cerr << *format++;
        cerr << '=' << f << ", ";
        _dbg(format + 1, r...);
    }
      
    template<typename T>
    ostream &operator<<(ostream& os, vector<T> V) {
        os << "[ "; for (auto v : V) os << v << ","; return os << " ]";
    }
   
    template<typename T>
    ostream &operator<<(ostream& os, set<T> V) {
        os << "[ "; for (auto v : V) os << v << ","; return os << " ]";
    }

    template<typename T>
    ostream &operator<<(ostream& os, multiset<T> V) {
        os << "[ "; for (auto v : V) os << v << ","; return os << " ]";
    }
 
    template<typename T1, typename T2>
    ostream &operator<<(ostream& os, map<T1, T2> V) {
        os << "[ "; for (auto v : V) os << v << ","; return os << " ]";
    }
  
    template<typename L, typename R>
    ostream &operator<<(ostream &os, pair<L, R> P) {
        return os << "(" << P.first << "," << P.second << ")";
    }

    #ifdef BACKLIGHT
    #define debug(...) cerr << "33[31m" << "[" << __LINE__ << "] : "; _dbg(#__VA_ARGS__, __VA_ARGS__); cerr << "33[0m";
    // #define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__); 
    #else
    #define debug(...)
    #endif
}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return l + rng() % (r - l + 1); }

using namespace Backlight;
const int N = 105;
const int C = 10000;
const int K = 1e7 + 5;
const int MOD = 1e9 + 7;              // 998244353 1e9 + 7
const int INF = 0x3f3f3f3f;             // 1e9 + 7 0x3f3f3f3f
const ll LLINF = 0x3f3f3f3f3f3f3f3f;    // 1e18 + 9 0x3f3f3f3f3f3f3f3f
const double eps = 1e-8;

int n, a[N], b[N], dp[N][N * N];
void solve(int Case) { // printf("Case #%d: ", Case);
    read(n);
    FOR(i, 1, n) read(a[i], b[i]);

    memset(dp, -1, sizeof(dp));

    int sum = 0;
    FOR(i, 1, n) sum += b[i];

    dp[0][0] = 0;
    FOR(i, 1, n) {
        ROF(j, n, 1) {
            ROF(c, C, a[i]) {
                if (dp[j - 1][c - a[i]] != -1)
                    dp[j][c] = max(dp[j][c], dp[j - 1][c - a[i]] + b[i]);
            }
        }
    }

    FOR(k, 1, n) {
        double ans = 0, x;
        FOR(c, 0, C) {
            if (dp[k][c] != -1) {
                x = min(double(c), (sum + dp[k][c]) / 2.0);
                ans = max(ans, x);
            }
        }
        printf("%.12f ", ans);
    }
    puts("");
}

int main() {
#ifdef BACKLIGHT
    freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    auto begin = std::chrono::steady_clock::now();
#endif

    // ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T = 1;
    // read(T);
    for (int _ = 1; _ <= T; _++) solve(_);

#ifdef BACKLIGHT
    auto end = std::chrono::steady_clock::now();
    auto duration = std::chrono::duration_cast<std::chrono::milliseconds>(end - begin);
    cerr << "33[32mTime Elasped: " << duration.count() << " ms33[0m" << endl;
#endif
    return 0;
}