华为机试 购物单

简介

购物单
有点没有讲清楚, 对于编号没有讲清楚. 不过还是挺复杂的. 看题解是, 有约束的背包问题.

code

代码抄的, 不过有点理解了

#include<iostream>
#include<vector>
using namespace std;
int max(int m, int n)
{
    return m>n?m:n;
}
int dp[3200];
int main()
{
    int N,n,v,p,q;
    cin >> N >> n;
    N = N/10;
 
    int *ZJ_Pri = new int[n+1]();  int *ZJ_Imp = new int[n+1]();
    int *FJ1_Pri = new int[n+1](); int *FJ1_Imp = new int[n+1]();
    int *FJ2_Pri = new int[n+1](); int *FJ2_Imp = new int[n+1]();
 
    for(int i=1; i<=n; i++)
    {
        cin >> v >> p >> q;
        v = v / 10;
        if(q == 0)
        {
            ZJ_Pri[i] = v;
            ZJ_Imp[i] = v * p;
        }
        else if(FJ1_Pri[q] == 0)
        {
            FJ1_Pri[q] = v;
            FJ1_Imp[q] = v * p;
        }
        else
        {
            FJ2_Pri[q] = v;
            FJ2_Imp[q] = v * p;
        }
    }
    for(int i = 1; i <= n; i++)//i---前i个物品
    {
        for(int j = N; j >=1; j--)//j--当前的钱数
        {
            if(j >= ZJ_Pri[i])
                dp[j] = max(dp[j], dp[ j-ZJ_Pri[i] ] + ZJ_Imp[i]);
            if(j >= ZJ_Pri[i] + FJ1_Pri[i])
                dp[j] = max(dp[j], dp[ j-ZJ_Pri[i]-FJ1_Pri[i] ] + ZJ_Imp[i] + FJ1_Imp[i]);
            if(j >= ZJ_Pri[i] + FJ2_Pri[i])
                dp[j] = max(dp[j], dp[ j-ZJ_Pri[i]-FJ2_Pri[i] ] + ZJ_Imp[i] + FJ2_Imp[i]);
            if(j >= ZJ_Pri[i] + FJ1_Pri[i] + FJ2_Pri[i])
                dp[j] = max(dp[j], dp[ j-ZJ_Pri[i]-FJ1_Pri[i]-FJ2_Pri[i] ] + ZJ_Imp[i] + FJ1_Imp[i] + FJ2_Imp[i]);
        }
    }
    cout << dp[N]*10 << endl;
 
    delete[] ZJ_Pri;
    delete[] ZJ_Imp;
    delete[] FJ1_Pri;
    delete[] FJ1_Imp;
    delete[] FJ2_Pri;
    delete[] FJ2_Imp;
 
    return 0;
}