算法课作业 3-14 最少费用购物问题 DP

给定物品种类数n(0到5),接下来n行每行3个数商品编码c_i(1到999),需要购买的数量k_i(1到5),单价p_i(1到999).
然后给定m(0到99)种优惠,接下来m行,每行第一个数字q表示这种优惠需要购买的物品种类数,接着q个数对(c,k)表示编码为c的商品买k个,每行最后一个数字表示买这些物品组合的优惠价.
输出购买所需物品的最少价格.

考虑dp
状态表示: dp[n1][n2][n3][n4][n5] d p [ n 1 ] [ n 2 ] [ n 3 ] [ n 4 ] [ n 5 ] 表示第1,2,3,4,5种物品购买数量分别为n_1,n_2,n_3,n_4,n_5的最小价格.
边界状态: dp[0][0][0][0][0]=0 d p [ 0 ] [ 0 ] [ 0 ] [ 0 ] [ 0 ] = 0
状态转移:设p_i为第i种优惠,将单价也视为一种优惠.
dp[n1][n2][n3][n4][n5]=minmi=1{dp[n1−pi1][n2−pi2][n3−pi3][n4−pi4][n5−pi5]} d p [ n 1 ] [ n 2 ] [ n 3 ] [ n 4 ] [ n 5 ] = m i n i = 1 m { d p [ n 1 − p i 1 ] [ n 2 − p i 2 ] [ n 3 − p i 3 ] [ n 4 − p i 4 ] [ n 5 − p i 5 ] }

记忆化搜索.

真是神奇,记得有时间补了电梯那道题,记忆化搜索之王.

/* LittleFall : Hello! */
#include 
using namespace std;typedef long long ll;
inline int read();inline void write(int x);
const int M = 5;

int mp[M], need[M]; //商品编码映射,所需商品个数
int dp[M][M][M][M][M];
struct Sale
{
    int num[M];
    int price = 0;
} sale[M]; //优惠
int sales; //优惠数

int mymin(int a, int b){return a == -1 ? b : min(a, b);}
int search()
{
    int &ans = dp[need[0]][need[1]][need[2]][need[3]][need[4]];
    if(ans != -1) return ans;

    for(int i = 0; i < sales; i++)
    {
        int flag = 1;
        for(int j = 0; j < M; j++)
            if(need[j] < sale[i].num[j])
                flag = 0, j = M;
        if(!flag) continue;
        for(int j = 0; j < M; j++) need[j] -= sale[i].num[j];
        ans = mymin(ans, search() + sale[i].price);
        for(int j = 0; j < M; j++) need[j] += sale[i].num[j];
    }
    return ans;
}
int main(void)
{
    freopen("3-14.txt", "r", stdin);

    int n = read();
    for(sales = 0; sales < n; sales++)
    {
        mp[read()] = sales;
        need[sales] = read();
        sale[sales].num[sales] = 1;
        sale[sales].price = read();
    }
    int m = read();
    for(; sales < n + m; sales++)
    {
        int k = read();
        while(k--)
        {
            int id = mp[read()];
            sale[sales].num[id] = read();
        }
        sale[sales].price = read();
    }
    memset(dp, -1, sizeof(dp));
    dp[0][0][0][0][0] = 0;
    write(search());

    return 0;
}

inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void write(int x)
{
     if(x<0) putchar('-'),x=-x;
     if(x>9) write(x/10);
     putchar(x%10+'0');
}