蓝桥杯算法提高VIP-促销购物 （C++代码）

解题思路:

注意事项:

参考代码:

#include <iostream>
#include <stdio.h>
#include <string.h>
#define INF 0x3f3f3f3f
#define min(a,b) (a)>(b)?(b):(a)
using namespace std;
 
struct Data1
{
    int num[1010], money;
}a[110];
 
struct Data2
{
    int c, k, p;
}b[10];
 
int n, m, res[6][6][6][6][6];
 
int main()
{
    scanf("%d", &n);
    int i, j;
    for (i = 0; i < n; i++)
    {
        int t;
        scanf("%d", &t);
        for (j = 0; j < t; j++)
        {
            int k ,l;
            scanf("%d%d", &k, &l);
            a[i].num[k] = l;
        }
        scanf("%d", &a[i].money);
    }
    scanf("%d", &m);
    for (i = 0;  i < m; i++)
    {
        scanf("%d%d%d", &b[i].c, &b[i].k, &b[i].p);
        a[n].money = b[i].p;            //将单价也加入优惠数组，只不过单价的话数量为1，只有一个物品，并且价格就是单价
        a[n++].num[b[i].c] = 1;
    }
   
    memset(res, INF, sizeof(res));
 
    res[0][0][0][0][0] = 0;
    int b1, b2,b3,b4,b5,m1,m2,m3,m4,m5;
    for (i = 0; i < n; i++)
    {
        m1 = a[i].num[b[0].c];
        m2 = a[i].num[b[1].c];
        m3 = a[i].num[b[2].c];
        m4 = a[i].num[b[3].c];
        m5 = a[i].num[b[4].c];            //每个优惠政策的每件物品的基本数量，在这个基本数量上进行dp讨论
        for (b1=m1; b1 <= b[0].k; b1++)
            for(b2 = m2; b2 <= b[1].k; b2++)
                for (b3 = m3; b3 <= b[2].k; b3++)
                    for(b4 = m4; b4 <= b[3].k; b4++)
                        for(b5 = m5; b5 <= b[4].k; b5++)
                            res[b1][b2][b3][b4][b5] = min(res[b1][b2][b3][b4][b5], res[b1-m1][b2-m2][b3-m3][b4-m4][b5-m5] + a[i].money);
 
    }
    printf("%d\n", res[b[0].k][b[1].k][b[2].k][b[3].k][b[4].k]);
    
    return 0;
}