PTA数据结构与算法题目集(中文) 7-39魔法优惠券 (25 分)

PTA数据结构与算法题目集(中文)  7-39魔法优惠券 (25 分)

7-39 魔法优惠券 (25 分)

 

在火星上有个魔法商店，提供魔法优惠券。每个优惠劵上印有一个整数面值K，表示若你在购买某商品时使用这张优惠劵，可以得到K倍该商品价值的回报！该商店还免费赠送一些有价值的商品，但是如果你在领取免费赠品的时候使用面值为正的优惠劵，则必须倒贴给商店K倍该商品价值的金额…… 但是不要紧，还有面值为负的优惠劵可以用！（真是神奇的火星）

例如，给定一组优惠劵，面值分别为1、2、4、-1；对应一组商品，价值为火星币M$7、6、-2、-3，其中负的价值表示该商品是免费赠品。我们可以将优惠劵3用在商品1上，得到M$28的回报；优惠劵2用在商品2上，得到M$12的回报；优惠劵4用在商品4上，得到M$3的回报。但是如果一不小心把优惠劵3用在商品4上，你必须倒贴给商店M$12。同样，当你一不小心把优惠劵4用在商品1上，你必须倒贴给商店M$7。

规定每张优惠券和每件商品都只能最多被使用一次，求你可以得到的最大回报。

输入格式:

输入有两行。第一行首先给出优惠劵的个数N，随后给出N个优惠劵的整数面值。第二行首先给出商品的个数M，随后给出M个商品的整数价值。N和M在[1, 1]之间，所有的数据大小不超过2​30​​，数字间以空格分隔。

输出格式:

输出可以得到的最大回报。

输入样例:

4 1 2 4 -1
4 7 6 -2 -3

输出样例:

43
题目分析：用优先队列来每次找出最好的情况，看了大佬的做法，其实只需要排序就好了，每次找出符号相同的最好情况 但这题我最后一个测试题无法通过

  1 #define _CRT_SECURE_NO_WARNINGS
  2 #include<stdio.h>
  3 #include<string.h>
  4 #include<malloc.h>
  5 #define INIFITY -65535
  6 typedef struct HeapStruct* H;
  7 struct HeapStruct
  8 {
  9     int Heap[1000010];
 10     int Size;
 11 }MaxHeap1,MaxHeap2;
 12 
 13 int Delete(H MaxHeap)
 14 {
 15     int Max = MaxHeap->Heap[1];
 16     int Tmp = MaxHeap->Heap[MaxHeap->Size--];
 17     int Parent, Child;
 18     for (Parent = 1; Parent * 2 <= MaxHeap->Size; Parent = Child)
 19     {
 20         Child = Parent * 2;
 21         if (Child != MaxHeap->Size && MaxHeap->Heap[Child] < MaxHeap->Heap[Child + 1])
 22             Child++;
 23         if (Tmp >= MaxHeap->Heap[Child])break;
 24         else
 25             MaxHeap->Heap[Parent] = MaxHeap->Heap[Child];
 26     }
 27     MaxHeap->Heap[Parent] = Tmp;
 28     return Max;
 29 }
 30 int DeleteM(H MinHeap)
 31 {
 32     int Min = MinHeap->Heap[1];
 33     int Tmp = MinHeap->Heap[MinHeap->Size--];
 34     int Parent, Child;
 35     for (Parent = 1; Parent * 2 <= MinHeap->Size; Parent = Child)
 36     {
 37         Child = Parent * 2;
 38         if (Child != MinHeap->Size && MinHeap->Heap[Child]>MinHeap->Heap[Child + 1])
 39             Child++;
 40         if (Tmp <= MinHeap->Heap[Child])break;
 41         else
 42             MinHeap->Heap[Parent] = MinHeap->Heap[Child];
 43     }
 44     MinHeap->Heap[Parent] = Tmp;
 45     return Min;
 46 }
 47 
 48 void PrecDown(H MaxHeap, int i)
 49 {
 50     int Parent, Child;
 51     int Tmp = MaxHeap->Heap[i];
 52     for (Parent = i; Parent * 2 <= MaxHeap->Size; Parent = Child)
 53     {
 54         Child = Parent * 2;
 55         if (Child != MaxHeap->Size && MaxHeap->Heap[Child] < MaxHeap->Heap[Child + 1])
 56             Child++;
 57         if (Tmp >= MaxHeap->Heap[Child])break;
 58         else
 59             MaxHeap->Heap[Parent] = MaxHeap->Heap[Child];
 60     }
 61     MaxHeap->Heap[Parent] = Tmp;
 62 }
 63 void BuildHeap(H MaxHeap)
 64 {
 65     for (int i = MaxHeap->Size / 2; i; i /= 2)
 66         PrecDown(MaxHeap, i);
 67 }
 68 void PrecDownM(H MinHeap, int i)
 69 {
 70     int Parent, Child;
 71     int Tmp = MinHeap->Heap[i];
 72     for (Parent = i; Parent * 2 <= MinHeap->Size; Parent = Child)
 73     {
 74         Child = Parent * 2;
 75         if (Child != MinHeap->Size && MinHeap->Heap[Child] > MinHeap->Heap[Child + 1])
 76             Child++;
 77         if (Tmp <= MinHeap->Heap[Child])break;
 78         else
 79             MinHeap->Heap[Parent] = MinHeap->Heap[Child];
 80     }
 81     MinHeap->Heap[Parent] = Tmp;
 82 }
 83 void BuildHeapM(H MinHeap)
 84 {
 85     for (int i = MinHeap->Size / 2; i; i /= 2)
 86         PrecDownM(MinHeap, i);
 87 }
 88 int main()
 89 {
 90     MaxHeap1.Heap[0] = MaxHeap2.Heap[0] = INIFITY;
 91     int N, M;
 92     int num,sum;
 93     scanf("%d", &N);
 94     MaxHeap1.Size = N;
 95     for (int i = 1; i <=MaxHeap1.Size; i++)
 96     {
 97         scanf("%d", &num);
 98         MaxHeap1.Heap[i] = num;
 99     }
100     scanf("%d", &M);
101     MaxHeap2.Size = M;
102     for (int i = 1; i <=MaxHeap2.Size; i++)
103     {
104         scanf("%d", &num);
105         MaxHeap2.Heap[i] = num;
106     }
107     BuildHeap(&MaxHeap1);
108     BuildHeap(&MaxHeap2);
109     sum = 0;
110     int num1, num2;
111     int t = N > M ? M : N;
112     while (MaxHeap1.Size!=0&&MaxHeap2.Size!=0)
113     {    
114         if (MaxHeap1.Heap[1] > 0 && MaxHeap2.Heap[1] > 0)
115         {
116             num1 = Delete(&MaxHeap1);
117             num2 = Delete(&MaxHeap2);
118             sum += num1 * num2;
119         }
120         else
121             break;
122     }
123     BuildHeapM(&MaxHeap1);
124     BuildHeapM(&MaxHeap2);
125     while (MaxHeap1.Size!=0&&MaxHeap2.Size!=0)
126     {
127         if (MaxHeap1.Heap[1] < 0 && MaxHeap2.Heap[1] < 0)
128         {
129             num1 = DeleteM(&MaxHeap1);
130             num2 = DeleteM(&MaxHeap2);
131             sum += num1 * num2;
132         }
133         else
134             break;
135     }
136     printf("%d", sum);
137 }

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