1 条题解

  • 0
    @ 2026-10-27 0:36:27

    思路与算法

    把每家公司分配0..M台看成一组选项,dp[i][j]记录前i家公司最多使用j台的最佳利润,枚举本公司台数并保存回溯决策。

    正确性说明

    对最后一家公司的台数分类,余下最优值必须由前i-1家公司取得;每种合法分配恰归入一个类别,取最大给出最优值。逐层保存决策能构造台数总和不超过M的同值方案。

    复杂度

    O(NM²D)时间、O(NMD)数位空间,D为利润数字长度;含二维表和方案初始化。

    易错点

    最多M台而非强制用完;0台利润为0,允许闲置。原样例//后的中文仅为解释,机器输入不包含解释。未给利润位数上限,代码内置十进制整数运算。

    参考代码(C++11)

    #include <bits/stdc++.h>
    using namespace std;
    struct cpp_int {
        string s;
        bool neg;
        cpp_int(long long v = 0) : s(to_string(v)), neg(v < 0) {
            if (neg)
                s.erase(s.begin());
        }
        void norm() {
            auto p = s.find_first_not_of('0');
            s = p == string::npos ? "0" : s.substr(p);
            if (s == "0")
                neg = false;
        }
        static int cmpabs(const cpp_int &a, const cpp_int &b) {
            if (a.s.size() != b.s.size())
                return a.s.size() < b.s.size() ? -1 : 1;
            return a.s == b.s ? 0 : a.s < b.s ? -1 : 1;
        }
        friend bool operator<(const cpp_int &a, const cpp_int &b) {
            if (a.neg != b.neg)
                return a.neg;
            int c = cmpabs(a, b);
            return a.neg ? c > 0 : c < 0;
        }
        friend bool operator==(const cpp_int &a, const cpp_int &b) {
            return a.neg == b.neg && a.s == b.s;
        }
        friend bool operator!=(const cpp_int &a, const cpp_int &b) {
            return !(a == b);
        }
        friend bool operator>(const cpp_int &a, const cpp_int &b) {
            return b < a;
        }
        friend bool operator<=(const cpp_int &a, const cpp_int &b) {
            return !(b < a);
        }
        friend bool operator>=(const cpp_int &a, const cpp_int &b) {
            return !(a < b);
        }
        cpp_int operator-() const {
            cpp_int r = *this;
            if (r.s != "0")
                r.neg = !r.neg;
            return r;
        }
        friend cpp_int operator+(const cpp_int &a, const cpp_int &b) {
            cpp_int r;
            r.s.clear();
            int carry = 0;
            if (a.neg == b.neg) {
                int i = (int)a.s.size() - 1, j = (int)b.s.size() - 1;
                while (i >= 0 || j >= 0 || carry) {
                    int v = carry + (i >= 0 ? a.s[i--] - '0' : 0) + (j >= 0 ? b.s[j--] - '0' : 0);
                    r.s += char('0' + v % 10);
                    carry = v / 10;
                }
                r.neg = a.neg;
            } else {
                const cpp_int *x = &a, *y = &b;
                if (cmpabs(a, b) < 0)
                    swap(x, y);
                int i = (int)x->s.size() - 1, j = (int)y->s.size() - 1;
                while (i >= 0) {
                    int v = x->s[i--] - '0' - carry - (j >= 0 ? y->s[j--] - '0' : 0);
                    carry = v < 0;
                    if (carry)
                        v += 10;
                    r.s += char('0' + v);
                }
                r.neg = x->neg;
            }
            reverse(r.s.begin(), r.s.end());
            r.norm();
            return r;
        }
        friend cpp_int operator-(const cpp_int &a, const cpp_int &b) {
            return a + (-b);
        }
        friend cpp_int operator*(const cpp_int &a, int b) {
            cpp_int r;
            r.s.clear();
            long long carry = 0;
            long long z = b;
            if (z < 0)
                z = -z;
            for (int i = (int)a.s.size() - 1; i >= 0; i--) {
                carry += (a.s[i] - '0') * z;
                r.s += char('0' + carry % 10);
                carry /= 10;
            }
            while (carry) {
                r.s += char('0' + carry % 10);
                carry /= 10;
            }
            reverse(r.s.begin(), r.s.end());
            r.neg = a.neg ^ (b < 0);
            r.norm();
            return r;
        }
        friend cpp_int operator/(const cpp_int &a, int b) {
            cpp_int r;
            r.s.clear();
            long long rem = 0, z = b;
            if (z < 0)
                z = -z;
            for (char c : a.s) {
                rem = rem * 10 + c - '0';
                r.s += char('0' + rem / z);
                rem %= z;
            }
            r.neg = a.neg ^ (b < 0);
            r.norm();
            return r;
        }
        cpp_int &operator+=(const cpp_int &b) {
            *this = *this + b;
            return *this;
        }
        cpp_int &operator-=(const cpp_int &b) {
            *this = *this - b;
            return *this;
        }
        cpp_int &operator*=(int b) {
            *this = *this * b;
            return *this;
        }
        friend istream &operator>>(istream &in, cpp_int &a) {
            string v;
            if (in >> v) {
                a.neg = v[0] == '-';
                if (v[0] == '-' || v[0] == '+')
                    v.erase(v.begin());
                a.s = v;
                a.norm();
            }
            return in;
        }
        friend ostream &operator<<(ostream &out, const cpp_int &a) {
            if (a.neg)
                out << '-';
            return out << a.s;
        }
    };
    
    struct DS {
        vector<int> p, s;
        DS(int n) : p(n), s(n, 1) {
            iota(p.begin(), p.end(), 0);
        }
        int find(int u) {
            while (p[u] != u) {
                p[u] = p[p[u]];
                u = p[u];
            }
            return u;
        }
        bool join(int a, int b) {
            a = find(a);
            b = find(b);
            if (a == b)
                return false;
            if (s[a] > s[b])
                swap(a, b);
            p[a] = b;
            s[b] += s[a];
            return true;
        }
    };
    
    int main() {
        int n, m;
        cin >> n >> m;
        vector<vector<cpp_int>> a(n + 1, vector<cpp_int>(m + 1)), d(n + 1, vector<cpp_int>(m + 1));
        vector<vector<int>> take(n + 1, vector<int>(m + 1));
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                cin >> a[i][j];
        for (int i = 1; i <= n; i++)
            for (int j = 0; j <= m; j++)
                for (int k = 0; k <= j; k++)
                    if (d[i - 1][j - k] + a[i][k] > d[i][j]) {
                        d[i][j] = d[i - 1][j - k] + a[i][k];
                        take[i][j] = k;
                    }
        cout << d[n][m] << "\n";
        vector<int> x(n + 1);
        for (int i = n, j = m; i; i--) {
            x[i] = take[i][j];
            j -= x[i];
        }
        for (int i = 1; i <= n; i++)
            cout << i << " " << x[i] << "\n";
    }
    
    • 1

    信息

    ID
    1945
    时间
    1000ms
    内存
    64MiB
    难度
    普及−
    标签
    递交数
    0
    已通过
    0
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