/* 
算法设计与分析作业 
    ---设备最优更新策略 
*/ 
#include<iostream>

void accept(int **a, int years); 
int max(int a, int b) { return  a > b ? a : b; } 
int main() { 
    using std::cin; 
    using std::cout; 
    using std::endl;

    cout << "请输入决策年数（i）：" << endl; 
    int temp; 
    while (!(cin >> temp)||temp < 0) { 
        cout << "请输入一个正整数：" << endl; 
        cin.clear(); 
        while (cin.get() != ‘\n‘); 
    } 
    int years = temp + 1; 
    
    //第i年买的设备在第t年能产生的利润 
    int **exProfit = new int*[years];//0-n年买的设备 
    for (int i = 0; i < years; i++) { 
        exProfit[i] = new int[years];//0-n年买的设备在第i年的利润 
    } 
    cout << "请依次输入利润：" << endl; 
    accept(exProfit, years); 
    //第i年买的设备在第t年产生的维修费用 
    int **fix = new int *[years]; 
    for (int i = 0; i < years; i++){ 
        fix[i] = new int[years]; 
    } 
    cout << "请依次输入维修费用：" << endl; 
    accept(fix, years); 
    //第i年买的设备在第t年更新时的费用 
    int **update = new int *[years]; 
    for (int i = 0; i < years; i++){ 
        update[i] = new int[years]; 
    } 
    cout << "请依次输入更新费用：" << endl; 
    accept(update, years); 
    //最优更新策略，true代表更新，false代表继续使用 
    bool **choice = new bool *[years]; 
    for (int i = 0; i < years; i++) { 
        choice[i] = new bool[years]; 
    } 
    for (int i = 0; i < years; i++){ 
        for (int j = 0; j < years; j++) 
            choice[i][j] = false; 
    } 
    //最终利润，决策年数之后的一年利润全部置0 
    int **profit = new int*[years]; 
    for (int i = 0; i < years; i++) { 
        profit[i] = new int[years + 1]; 
    } 
    for (int i = 0; i < years; i++) 
    { 
        for (int j = 0; j < years + 1; j++) 
            profit[i][j] = 0; 
    } 
    //倒序进行计算，第i年，可以使用的设备要么就是之前一直没更新，要么就是第j年买的设备 
    //例如，第三年，设备的状态就有三种选择，一是使用第0年的设备，二是在第一年进行了更新，三是第二年进行了更新 
    //分别对每个状态进行选择更新或者继续使用 
    //第i年，有两种选择，要么更新，要么继续使用 
    //分别计算更新和继续使用的利润，选择两者中比较大的。 
    for (int  i = temp; i >0 ; i--) 
    { 
        for (int j = 0; j < i; j++) 
        { 
            int keep = exProfit[j][i] - fix[j][i] + profit[j][i + 1]; 
            int refresh = exProfit[i][i] - fix[i][i] - update[j][i] + profit[i][i + 1]; 
            profit[j][i] = max(keep, refresh); 
            if (keep < refresh) { 
                choice[i][j] = true; 
            } 
        } 
    } 
    int j = 0; 
    for (int i = 1; i < years; i++) 
    { 
        if (choice[i][j]) { 
            j = i; 
            std::cout <<"第"<<i<< "年更新"<<std::endl; 
            continue; 
        } 
            std::cout << "第" << i << "年继续使用" << std::endl;    
    } 
    cout << "利润最大为：" << profit[0][1]; 
    delete[]profit; 
    for (int i = 0; i < years; i++) { 
        delete[] fix[i]; 
    } 
    delete[]fix; 
    for (int i = 0; i < years; i++) { 
        delete[] exProfit[i]; 
    } 
    delete[]exProfit; 
    for (int i = 0; i < years; i++) { 
        delete[] update[i]; 
    } 
    delete[]update; 
    for (int i = 0; i < years; i++) { 
        delete[] choice[i]; 
    } 
    delete[]choice; 
}

void accept(int **a ,int years) { 
    for (int i = 0; i < years; i++) { 
        std::cout << "第"<<i<<"年："; 
        int j = 0; 
        a[i][j] = 0; 
        j++; 
        while (j < years) 
        { 
            if (j < i) 
                a[i][j] = 0; 
            else { 
                while (!(std::cin >> a[i][j])||a[i][j] < 0) { 
                    if(i == 0) 
                        std::cout << "请从第" <<j<<"个数重新输入"<< std::endl; 
                    else 
                        std::cout << "请从第" << j-i+1 << "个数重新输入" << std::endl; 
                    std::cout << "请输入正整数"; 
                    std::cin.clear(); 
                    while (std::cin.get() != ‘\n‘); 
                } 
            } 
            j++; 
        } 
    } 
}