codeforces Round #413 Div. 1 + Div. 2 B T-shirt buying
来源:互联网 发布:c语言怎么解析 编辑:程序博客网 时间:2024/05/22 08:00
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.
5300 200 400 500 9111 2 1 2 32 1 3 2 162 3 1 2 1 1
200 400 300 500 911 -1
21000000000 11 11 222 1
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;}
- Codeforces Round #413 Div. 1 + Div. 2 B. T-shirt buying
- Codeforces Round #413 (Div.1 + Div. 2) B. T-shirt buying
- codeforces Round #413 Div. 1 + Div. 2 B T-shirt buying
- Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2) B. T-shirt buying
- codeforcesPlayrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2)B题B. T-shirt buying
- Codeforces Round #413 B. T-shirt buying
- Codeforces Round #413 B. T-shirt buying
- codeforces round #412 B. T-shirt buying
- CodeForces 799B (B) T-shirt buying
- codeforces #413 B T-shirt buying(set快速查找)
- Codeforces 799B T-shirt buying 题解
- Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3) B. T-Shirt Hunt(模拟)
- (Codeforces Round #413, rated, Div. 1 + Div. 2)(A+B)
- Codeforces Round #142 (Div. 2) B - T-primes
- Codeforces Round #408 (Div. 2) A Buying A House
- codeforces round#408(div.2) A. Buying A House
- Codeforces Round #408 (Div. 2) A. Buying A House
- Codeforces Round #413(Div. 1 + Div. 2)
- BZOJ 2818: Gcd 线形筛
- 线性查找法————使用查找键逐个与数组元素进行比较实现查找
- C++的学习过程
- thinkphp 错误页面404定制
- matlab+数字图像处理教程之-基本概念
- codeforces Round #413 Div. 1 + Div. 2 B T-shirt buying
- h5
- 解析js中的原型链和在原型链中对象属性查找
- 友盟第三方QQ简版快速上手步骤
- Java数组和字符串的长度计算
- 安装jupyter,一款python的web编辑器
- 1090. Highest Price in Supply Chain (25)
- CSS并不简单--带你认识不一样的渐变
- java:volatile关键字解析