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.
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.
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.
5 300 200 400 500 911 1 2 1 2 3 2 1 3 2 1 6 2 3 1 2 1 1
200 400 300 500 911 -1
2 1000000000 1 1 1 1 2 2 2 1
1 1000000000
感觉应该有很多做法，我用的是优先队列，3个队列代表不同颜色，读入数据并处理后易知3个队列的大小之和为2*n，所以为了防止一件T桖多次卖出，用来vis标记衣服的序号。
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
int vis[200005];
int pr[200005];
struct node{
int p,no;
bool operator <(const node &a)const{
return p>a.p;
}
};
node t;
priority_queue<node> a;
priority_queue<node> b;
priority_queue<node> c;
int n,m,temp,flag;
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n;i++)
scanf("%d",&pr[i]);
for(int i=0;i<n;i++)
{
scanf("%d",&temp);
t.p=pr[i];
t.no=i;
if(temp==1)
a.push(t);
else if(temp==2)
b.push(t);
else
c.push(t);
}
for(int i=0;i<n;i++)
{
scanf("%d",&temp);
t.p=pr[i];
t.no=i;
if(temp==1)
a.push(t);
else if(temp==2)
b.push(t);
else
c.push(t);
}
memset(vis,0,sizeof(vis));
scanf("%d",&m);
while(m--)
{
scanf("%d",&temp);
if(temp==1)
{
flag=0;
while(!a.empty()&&flag==0)
{
t=a.top();
a.pop();
if(vis[t.no]==0)
{
printf("%d ",t.p);
flag=1;
vis[t.no]=1;
}
}
if(!flag)
printf("-1 ");
}
else if(temp==2)
{
flag=0;
while(!b.empty()&&flag==0)
{
t=b.top();
b.pop();
if(vis[t.no]==0)
{
printf("%d ",t.p);
flag=1;
vis[t.no]=1;
}
}
if(!flag)
printf("-1 ");
}
else
{
flag=0;
while(!c.empty()&&flag==0)
{
t=c.top();
c.pop();
if(vis[t.no]==0)
{
printf("%d ",t.p);
flag=1;
vis[t.no]=1;
}
}
if(!flag)
printf("-1 ");
}
}
printf("\n");
}
return 0;
}