3 seconds
256 megabytes
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 题目大意：
在一个服装店里，有n件T-shirt待售，给定他们的价格 以及每件衣服前面的颜色 和后面的颜色（只可能有三种颜色1/2/3）
现在有m名顾客轮流进入商店购买衣服，每个人都有自己喜欢的颜色（他们都希望买到自己喜欢的颜色的衣服 并且花费要尽可能的少）
输出每一名顾客的最少花费 如果找不到自己喜欢的衣服，他将不会购买任何东西（输出-1）
顾客一个一个的轮流进入服装店，服务员每次只会服务当前的顾客 解题思路：
刚开始写的时候没有考虑到n的范围，直接将价格存入数组然后sort排序之后暴力查找 果断TLE
之后又用vector（容器）优化了一丢丢，结果还是TLE
最后看了学长的代码发现他用了一个神奇的东西（STL里的set） 用它优化之后果断A了
利用set将各种颜色的衣服分类存入一起（它神奇的地方在于 再存的过程中能够自动的按照从小到大排序） 关于 set的具体用法请看这里 ：http://www.cnblogs.com/yoke/p/6867302.html
AC代码：
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <vector>
#include <set> using namespace std; int main()
{
set <int >vec[];
set<int>:: iterator it;
int n,m,a,b,c,i;
int p[];
while (~scanf("%d",&n))
{
for (i = ; i < n; i ++)
scanf("%d",&p[i]);
for (i = ; i < n; i ++){
scanf("%d",&a);
vec[a].insert(p[i]); // 将a颜色的T-shirt价格存到一起
}
for (i = ; i < n; i ++){
scanf("%d",&b); // 将b颜色的T-shirt价格存到一起
vec[b].insert(p[i]);
} scanf("%d",&m);
while (m --)
{
scanf("%d",&c);
if (vec[c].size() == ) // c颜色的T-shirt已经卖完了
printf("-1 ");
else
{
it = vec[c].begin();
int k = *it;
printf("%d ",k); it = vec[].find(k); // 将已经卖掉T-shirt清除
if (it != vec[].end())
vec[].erase(it);
it = vec[].find(k);
if (it != vec[].end())
vec[].erase(it);
it = vec[].find(k);
if (it != vec[].end())
vec[].erase(it);
}
}
printf("\n");
}
return ;
}
欢迎各位大佬指教！！！