USACO 3.3 Shopping Offers (shopping)

/*Main idearDP problem. Take each offer as a single choise for DP.dp[a1][a2][a3][a4][a5] = min{dp[a1-prodcut_num[i][1]][a2-prodcut_num[i][2]]                        [a1-prodcut_num[i][3]][a1-prodcut_num[i][4]][a1-prodcut_num[i][5]] + product_num[i][0]}ak - product_num[i][k] >= 0;  1 <= i <= offers_num;boundary conditions: dp[0][0][0][0][0] = 0;dp[a1][a2][a3][a4][a5] denote the least money we cost to buy a1 product 1, a2 product 2 .... a5 product 5;product[i][1] if i >= 1, it denote the number of product 1 in offer i; if i == 0; it denote the reduced priceof offer i;*//*Executing...   Test 1: TEST OK [0.000 secs, 3536 KB]   Test 2: TEST OK [0.000 secs, 3536 KB]   Test 3: TEST OK [0.000 secs, 3536 KB]   Test 4: TEST OK [0.011 secs, 3536 KB]   Test 5: TEST OK [0.011 secs, 3536 KB]   Test 6: TEST OK [0.000 secs, 3536 KB]   Test 7: TEST OK [0.000 secs, 3536 KB]   Test 8: TEST OK [0.000 secs, 3536 KB]   Test 9: TEST OK [0.011 secs, 3536 KB]   Test 10: TEST OK [0.022 secs, 3536 KB]   Test 11: TEST OK [0.032 secs, 3536 KB]   Test 12: TEST OK [0.032 secs, 3536 KB]All tests OK.*//*ID: haolink1PROG: shoppingLANG: C++*///#include <iostream>#include <fstream>#include <cstring>using namespace std;int dp[6][6][6][6][6];int product_num[105][6];int product_code[1000];//product_code[i] denote product i 's code by the order they are encountered;int ori_mark = 0;int buy_num[6];int offers_num = 0;ifstream fin("shopping.in");ofstream cout("shopping.out");inline int Min(int a,int b){    return a <= b ? a : b;}// Easy to code and only compute the necessary element of dp[]..[]; int Memorized_DP(int a1,int a2, int a3,int a4,int a5){    if(dp[a1][a2][a3][a4][a5] != ori_mark)        return dp[a1][a2][a3][a4][a5];    int min_price = ori_mark,w1,w2,w3,w4,w5;    for(int i = 1; i <= offers_num; i++){        w1 = a1 - product_num[i][1];        w2 = a2 - product_num[i][2];        w3 = a3 - product_num[i][3];        w4 = a4 - product_num[i][4];        w5 = a5 - product_num[i][5];        if(w1 < 0 || w2 < 0 || w3 < 0 || w4 < 0 || w5 < 0) continue;        if(dp[w1][w2][w3][w4][w5] == ori_mark)            dp[w1][w2][w3][w4][w5] = Memorized_DP(w1,w2,w3,w4,w5);          if(min_price > dp[w1][w2][w3][w4][w5]+product_num[i][0]){            min_price = dp[w1][w2][w3][w4][w5]+product_num[i][0];         }    }     return min_price; }int main(){    //init    int s = 0, num = 0, c = 0, k = 0, code = 0, b = 0;    fin >> s;     for(int i = 1; i <= s; i++){        fin >> num;        for(int j = 0; j < num; j++){            fin >> c >> k;            if(product_code[c] == 0) product_code[c] = ++code;            product_num[i][product_code[c]] = k;        }        fin >> product_num[i][0];    }    fin >> b;    offers_num = s + b;    //Single product can also be regarded as a offer.    for(int i = 1; i <= b; i++){        fin >> c >> k;        if(product_code[c] == 0) product_code[c] = ++code;        product_num[s+i][product_code[c]] = 1;        buy_num[product_code[c]] = k;        fin >> product_num[s+i][0];    }    //Note give each element of dp[][][][][] a big enough initial value     //except dp[0][0][0][0][0], because finally what we want is exactly     //a1,a2,a3,a4,a5 for dp[a1][a2][a3][a4][a5];    memset(dp,0xF,sizeof(dp));    ori_mark = dp[0][0][0][0][0];    dp[0][0][0][0][0] = 0;    //DP     cout << Memorized_DP(buy_num[1],buy_num[2],buy_num[3],buy_num[4],buy_num[5]) << endl;    return 0;}