codeforces Round #413 Div. 1 + Div. 2 B T-shirt buying

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本文介绍了一种模拟加贪心算法解决衬衫选购问题的方法。通过将衬衫按颜色分类并排序，每次为顾客选择最便宜且颜色符合要求的衬衫，有效解决了顾客选购问题。

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.

Examples

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

output
200 400 300 500 911 -1 

input
2
1000000000 1
1 1
1 2
2
2 1

output
1 1000000000 

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

模拟+贪心~

（其实我还是不知道这是什么~）

（代码看起来很长，其实就是ctrl+c/ctrl+v而已，但是这样容易出错需谨慎！）

衬衫显然不分前后，而且总共只有三种颜色，所以我们就能根据颜色把所有衬衫分成6种情况，用颜色的乘积代替，分别是1,4,9,2,3,6，那么我们把所有衬衫归类，排序，记录现在用了哪些衬衫，更新答案即可。其实和NOIP2016D2T2差不多。

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;

int n,m,a[200001],b[200001],c[200001],x,tot1,tot2,tot3,id,now1,now2,minn,now3,now,num[4][200001],noww[4];

struct node{
	int x,y,z;
}p[200001];

int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0' || ch>'9') {if(ch=='-') f=-1;ch=getchar();}
	while(ch>='0' && ch<='9') {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}

int main()
{
	n=read();
	for(int i=1;i<=n;i++) p[i].x=read();
	for(int i=1;i<=n;i++) p[i].y=read();
	for(int i=1;i<=n;i++) p[i].z=read();
	for(int i=1;i<=n;i++)
	{
		int kkz=p[i].y*p[i].z;
		if(p[i].y==p[i].z) num[p[i].y][++num[p[i].y][0]]=p[i].x;
		else if(kkz==2) a[++tot1]=p[i].x;
		else if(kkz==3) b[++tot2]=p[i].x;
		else c[++tot3]=p[i].x;
	}
	sort(a+1,a+tot1+1);
	sort(b+1,b+tot2+1);
	sort(c+1,c+tot3+1);
	for(int i=1;i<=3;i++) sort(num[i]+1,num[i]+num[i][0]+1); 
	m=read();
	while(m--)
	{
		now=0;x=read();minn=1000000001;id=0;
		if(x==1)
		{
			if(num[x][0]>noww[x])
			{
				now=1;
				if(num[x][noww[x]+1]<minn)
				{
					minn=num[x][noww[x]+1];
					id=1;
				}
			}
			if(tot1-now1)
			{
				now=1;
				if(a[now1+1]<minn)
				{
					minn=a[now1+1];
					id=2;
				}
			}
			if(tot2-now2)
			{
				now=1;
				if(b[now2+1]<minn)
				{
					minn=b[now2+1];
					id=3;
				}
			}
			if(!now) printf("-1 ");
			else
			{
				printf("%d ",minn);
				if(id==1) noww[x]++;
				else if(id==2) now1++;
				else now2++;
			}
		}
		else if(x==2)
		{
			if(num[x][0]>noww[x])
			{
				now=1;
				if(num[x][noww[x]+1]<minn)
				{
					minn=num[x][noww[x]+1];
					id=1;
				}
			}
			if(tot1-now1)
			{
				now=1;
				if(a[now1+1]<minn)
				{
					minn=a[now1+1];
					id=2;
				}
			}
			if(tot3-now3)
			{
				now=1;
				if(c[now3+1]<minn)
				{
					minn=c[now3+1];
					id=3;
				}
			}
			if(!now) printf("-1 ");
			else
			{
				printf("%d ",minn);
				if(id==1) noww[x]++;
				else if(id==2) now1++;
				else now3++;
			}
		}
		else
		{
			if(num[x][0]>noww[x])
			{
				now=1;
				if(num[x][noww[x]+1]<minn)
				{
					minn=num[x][noww[x]+1];
					id=1;
				}
			}
			if(tot3-now3)
			{
				now=1;
				if(c[now3+1]<minn)
				{
					minn=c[now3+1];
					id=2;
				}
			}
			if(tot2-now2)
			{
				now=1;
				if(b[now2+1]<minn)
				{
					minn=b[now2+1];
					id=3;
				}
			}
			if(!now) printf("-1 ");
			else
			{
				printf("%d ",minn);
				if(id==1) noww[x]++;
				else if(id==2) now3++;
				else now2++;
			}
		}
	}
	return 0;
}