题目如下

一包共 nn 件T恤被送到了商店。每件T恤被三个正整数 pipi​ , aiai​ 和 bibi​ 所描述。 pipi​ 是第 ii 件T恤的价格， aiai​ 是第 ii 件T恤正面的颜色， bibi​ 是第 ii 件T恤背面的颜色。所有 pipi​ 都各不相同， aiai​ 和 bibi​ 都是1~3的整数。

有 mm 个顾客，每人都仅想买一件T恤。第 ii 个顾客最喜欢的颜色是 cjcj​ 。

如果一件T恤至少一面（前或后）的颜色是顾客喜欢的，他才会将其买下；如果有多件T恤符合条件，他会选择最便宜的；如果没有符合条件的，他什么都不会买。我们假定顾客都乖巧地排着队一个一个来，前一个顾客走后第二个才能被接待。

你需要计算每个顾客分别花了多少钱。

输入格式：

 

The first line contains single integer nn ( 1n2000001<=n<=200000 ) — the number of t-shirts.

The following line contains sequence of integers p1p2...pnp1​,p2​,...,pn​ ( 1pi10000000001<=pi​<=1000000000 ), where pipi​equals to the price of the ii -th t-shirt.

The following line contains sequence of integers a1a2...ana1​,a2​,...,an​ ( 1ai31<=ai​<=3 ), where aiai​ equals to the front color of the ii -th t-shirt.

The following line contains sequence of integers b1b2...bnb1​,b2​,...,bn​ ( 1bi31<=bi​<=3 ), where bibi​ equals to the back color of the ii -th t-shirt.

The next line contains single integer mm ( 1m2000001<=m<=200000 ) — the number of buyers.

The following line contains sequence c1c2...cmc1​,c2​,...,cm​ ( 1cj31<=cj​<=3 ), where cjcj​ equals to the favorite color of the jj -th buyer. The buyers will come to the shop in the order they are given in the input. Each buyer is served only after the previous one is served.

 

输出格式：

 

Print to the first line mm integers — the jj -th integer should be equal to the price of the t-shirt which the jj -th buyer will buy. If the jj -th buyer won't buy anything, print -1.

 

 

　　这道题其实就是一个很简单的优先队列的调用。因为数据很大，单独循环会T。所以我们使用优先队列，并按照颜色的不同将结构体存入优先队列，每个人遍历一遍队列。由于价格从小到大，我们需要重载运算符。关于重载运算符的话，我此前学过很多次小詹的代码，但是总是记不住，今天试着用friend写一下，瞬间觉得充满了美感。

 

1 friend bool operator < (Node x,Node y)
2 
3 {
4 　　return x.minn < y.minn;
5 }

 

 

 

此题具体代码如下：

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<algorithm>
 4 #include<queue>
 5 using namespace std;
 6 typedef long long ll;
 7 int read()
 8 {
 9     int a = 0, b = 1;
10     char c = getchar();
11     while(c < '0' or c > '9')
12     {
13         if(c == '-') b = -1;
14         c = getchar();
15     }
16     while(c >= '0' and c <= '9')
17     {
18         a = a*10 + c - '0';
19         c = getchar();
20     }
21     return a*b;
22 }
23 ll a[100005],b[100005],p[100005],n,vis[100005],t,m;
24 struct node
25 {
26     int a,b,num,p;
27     bool operator < (const node & other)const
28     {
29         return p > other.p || (p == other.p && num > other.num);
30     }
31 }e[10000005];
32 int main()
33 {
34     priority_queue<node>k[4];
35     n = read();
36     for(int i=1; i<=n; i++) e[i].num = i;
37     for(int i=1; i<=n; i++) e[i].p = read();
38     for(int i=1; i<=n; i++) {e[i].a = read();k[e[i].a].push(e[i]);}
39     for(int i=1; i<=n; i++) {e[i].b = read();k[e[i].b].push(e[i]);}
40     m = read();
41     for(int i=1; i<=m; i++)
42     {
43         t = read();
44         if(k[t].empty())
45         {
46             printf("-1\n");
47             continue;
48         }
49         int c = k[t].top().num;
50         while(vis[c] == 1)
51         {
52             if(k[t].empty())
53             {
54                 break;
55             }
56             k[t].pop();
57             c = k[t].top().num;
58         }
59         if(!k[t].empty())
60         {printf("%d ",e[c].p);k[t].pop();}
61         else
62         printf("-1\n");
63         vis[c] = 1;
64     }
65     return 0;
66 }