Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2)B. T-shirt buying（想法题）

传送门

Description

A new pack of n t-shirts came to a shop. Each of the t-shirts is characterized by three integers pi, ai and bi, where pi is the price of the i-th t-shirt, ai is front color of the i-th t-shirt and bi is back color of the i-th t-shirt. All values pi are distinct, and values ai and bi are integers from 1 to 3.

m buyers will come to the shop. Each of them wants to buy exactly one t-shirt. For the j-th buyer we know his favorite color cj.

A buyer agrees to buy a t-shirt, if at least one side (front or back) is painted in his favorite color. Among all t-shirts that have colors acceptable to this buyer he will choose the cheapest one. If there are no such t-shirts, the buyer won't buy anything. Assume that the buyers come one by one, and each buyer is served only after the previous one is served.

You are to compute the prices each buyer will pay for t-shirts.

Input

The first line contains single integer n (1 ≤ n ≤ 200 000) — the number of t-shirts.

The following line contains sequence of integers p1, p2, ..., pn (1 ≤ pi ≤ 1 000 000 000), where pi equals to the price of the i-th t-shirt.

The following line contains sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 3), where ai equals to the front color of the i-th t-shirt.

The following line contains sequence of integers b1, b2, ..., bn (1 ≤ bi ≤ 3), where bi equals to the back color of the i-th t-shirt.

The next line contains single integer m (1 ≤ m ≤ 200 000) — the number of buyers.

The following line contains sequence c1, c2, ..., cm (1 ≤ cj ≤ 3), where cj equals to the favorite color of the j-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.

Output

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

Sample Input

5
300 200 400 500 911
1 2 1 2 3
2 1 3 2 1
6
2 3 1 2 1 1

2
1000000000 1
1 1
1 2
2
2 1

Sample Output

200 400 300 500 911 -1 

1 1000000000 

思路

题意：

有n件T-shirt，每件正反都有颜色，有m个顾客，希望买到的T-shirt至少有一面的颜色是自己喜欢的颜色，并且价格最低，依次输出m个顾客买到的T-shirt的号数，没有则输出-1.

题解:

 set维护颜色与价格的组合值。

 

+ View Code?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39

#include<bits/stdc++.h> using namespace std; const int maxn = 200005; int p[maxn],a[maxn],b[maxn]; bool vis[maxn]; set<pair<int,int> >s[5];   int main() {     int n,m,t;     scanf("%d",&n);     for (int i = 0;i < n;i++)    scanf("%d",&p[i]);     for (int i = 0;i < n;i++)    scanf("%d",&a[i]);     for (int i = 0;i < n;i++)    scanf("%d",&b[i]);     for (int i = 0;i < n;i++)     {         s[a[i]].insert(make_pair(p[i],i));         s[b[i]].insert(make_pair(p[i],i));         vis[i] = true;     }     scanf("%d",&m);     while (m--)     {         int ans = -1;         scanf("%d",&t);         while (!s[t].empty())         {             int pos = (*(s[t].begin())).second;             s[t].erase(s[t].begin());             if (!vis[pos])  continue;             ans = p[pos];             vis[pos] = false;             break;         }         printf("%d ",ans);     }     puts(" ");     return 0; }