hdu6438 Buy and Resell 买卖物品 ccpc网络赛 贪心
题目描述:
有n座城市,每座城市都可以对一个物品进行一次的买进或者卖出,可以同时拥有多个物品,计算利润最大值,并且交易次数要最少。(买入卖出算两次操作)
思路:
建立两个小根堆 优先队列,q1放可以买的物品,q2放可以卖的物品。
如果两个队列都是空的,则把这个物品放入q1.
如果q1是有的,而q2是空的,则把a[i]和q1的顶比一下,如果比他大,则q1 pop一次,把a[i]塞入q2,并且把差值累计到ans上。
如果q1无,q2有,则拿a[i]和q2顶比一下,如果比它大,则把q2顶元素放入q1,a[i]放入q2,差值累积到ans上。
如果两个都有,如果a[i]和两个队列顶部元素差值相同,则优先替换q2的,计算差值,a[i]放入q2,原来的q2顶放入q1.这样可以保证交易天数最小。 如果不相等,则替换放入大的那个,元素也要放到应该放的队列里,如果不能替换,就放入第一个。
这样做是因为我们只关心一个物品卖出后的利润,所以买入的价格不重要,重要的是卖出后的价格,因为这个是可以被替换的,而相同情况下,要减少天数,所以先替换已经卖出的东西。
感谢薛佬教我!
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string.h>
#include<sstream>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<bitset>
#define CLR(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
inline int rd(void) {
int x=0;
int f=1;
char s=getchar();
while(s<'0'||s>'9') {
if(s=='-')f=-1;
s=getchar();
}
while(s>='0'&&s<='9') {
x=x*10+s-'0';
s=getchar();
}
x*=f;
return x;
}
priority_queue<ll,vector<ll> ,greater<ll> >q1;
priority_queue<ll,vector<ll> ,greater<ll> >q2;
int n;
const int maxn=100010;
ll a[maxn];
int main() {
int T;
cin>>T;
while(T--) {
while(!q1.empty())q1.pop();
while(!q2.empty())q2.pop();
scanf("%d",&n);
ll sum=0;
for(int i=1; i<=n; i++) {
scanf("%lld",&a[i]);
if(q2.empty()) {
if(q1.empty()) {
q1.push(a[i]);
} else {
ll temp=q1.top();
if(a[i]>temp) {
sum+=a[i]-temp;
q1.pop();
q2.push(a[i]);
} else {
q1.push(a[i]);
}
}
} else {
if(!q1.empty()) {
ll temp1=q1.top();
ll temp2=q2.top();
if(a[i]>temp2&&temp1>=temp2) {
q1.push(temp2);
q2.pop();
q2.push(a[i]);
sum+=a[i]-temp2;
} else if(a[i]>temp1) {
q1.pop();
sum+=a[i]-temp1;
q2.push(a[i]);
} else {
q1.push(a[i]);
}
} else {
ll temp=q2.top();
if(a[i]>temp) {
sum+=a[i]-temp;
q2.pop();
q1.push(temp);
q2.push(a[i]);
} else
q1.push(a[i]);
}
}
}
printf("%lld %d\n",sum,q2.size()*2);
}
}
Buy and Resell
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 722 Accepted Submission(s): 188
Problem Description
The Power Cube is used as a stash of Exotic Power. There are n cities numbered 1,2,…,n where allowed to trade it. The trading price of the Power Cube in the i-th city is ai dollars per cube. Noswal is a foxy businessman and wants to quietly make a fortune by buying and reselling Power Cubes. To avoid being discovered by the police, Noswal will go to the i-th city and choose exactly one of the following three options on the i-th day:
1. spend ai dollars to buy a Power Cube
2. resell a Power Cube and get ai dollars if he has at least one Power Cube
3. do nothing
Obviously, Noswal can own more than one Power Cubes at the same time. After going to the n cities, he will go back home and stay away from the cops. He wants to know the maximum profit he can earn. In the meanwhile, to lower the risks, he wants to minimize the times of trading (include buy and sell) to get the maximum profit. Noswal is a foxy and successful businessman so you can assume that he has infinity money at the beginning.
Input
There are multiple test cases. The first line of input contains a positive integer T (T≤250), indicating the number of test cases. For each test case:
The first line has an integer n. (1≤n≤105)
The second line has n integers a1,a2,…,an where ai means the trading price (buy or sell) of the Power Cube in the i-th city. (1≤ai≤109)
It is guaranteed that the sum of all n is no more than 5×105.
Output
For each case, print one line with two integers —— the maximum profit and the minimum times of trading to get the maximum profit.
Sample Input
3 4 1 2 10 9 5 9 5 9 10 5 2 2 1
Sample Output
16 4 5 2 0 0
Hint
In the first case, he will buy in 1, 2 and resell in 3, 4. profit = - 1 - 2 + 10 + 9 = 16 In the second case, he will buy in 2 and resell in 4. profit = - 5 + 10 = 5 In the third case, he will do nothing and earn nothing. profit = 0
hdu6438 Buy and Resell 买卖物品 ccpc网络赛 贪心的更多相关文章
- HDU 6438 Buy and Resell ( 2018 CCPC 网络赛 && 贪心 )
题目链接 题意 : 给出一些数.你可以从左到右对这些数进行三种操作花费 Ai 买入东西.以 Ai 价格卖出你当前有的东西.或者什么都不做.现在问你可以获取的最大利益是多少? 分析 : 和 CF 867 ...
- 2018 CCPC网络赛
2018 CCPC网络赛 Buy and Resell 题目描述:有一种物品,在\(n\)个地点的价格为\(a_i\),现在一次经过这\(n\)个地点,在每个地点可以买一个这样的物品,也可以卖出一个物 ...
- 树形DP CCPC网络赛 HDU5834 Magic boy Bi Luo with his excited tree
// 树形DP CCPC网络赛 HDU5834 Magic boy Bi Luo with his excited tree // 题意:n个点的树,每个节点有权值为正,只能用一次,每条边有负权,可以 ...
- (四面体)CCPC网络赛 HDU5839 Special Tetrahedron
CCPC网络赛 HDU5839 Special Tetrahedron 题意:n个点,选四个出来组成四面体,要符合四面体至少四条边相等,若四条边相等则剩下两条边不相邻,求个数 思路:枚举四面体上一条线 ...
- ccpc 网络赛 hdu 6155
# ccpc 网络赛 hdu 6155(矩阵乘法 + 线段树) 题意: 给出 01 串,要么询问某个区间内不同的 01 子序列数量,要么把区间翻转. 叉姐的题解: 先考虑怎么算 \(s_1, s_2, ...
- 2018中国大学生程序设计竞赛 - 网络选拔赛 hdu6438 Buy and Resell 买入卖出问题 贪心
Buy and Resell Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)To ...
- 2018中国大学生程序设计竞赛 - 网络选拔赛 1001 - Buy and Resell 【优先队列维护最小堆+贪心】
题目传送门:http://acm.hdu.edu.cn/showproblem.php?pid=6438 Buy and Resell Time Limit: 2000/1000 MS (Java/O ...
- 2018年 CCPC 网络赛 赛后总结
历程:由于只是网络赛,所以今天就三开了.一开始的看题我看了d题,zz和jsw从头尾看起来,发现c题似乎可做,和费马大定理有关,于是和zz一起马上找如何计算勾股数的方法,比较慢的A掉了,而jsw此时看了 ...
- 2019杭电多校&CCPC网络赛&大一总结
多校结束了, 网络赛结束了.发现自己还是太菜了,多校基本就是爆零和签到徘徊,第一次打这种高强度的比赛, 全英文,知识点又很广,充分暴露了自己菜的事实,发现数学还是很重要的.还是要多刷题,少玩游戏. 网 ...
随机推荐
- java判断姓是否合格 百家姓
package util; import java.lang.reflect.Array; public class FirstName { public static boolean ClearNa ...
- OpenCV笔记 1
Structure Contains Represents CvPoint int x, y Point in image CvPoint2D32f float x, y Points in ...
- isinstance判断某个对象是否是某个类创建的
#!/usr/bin/env python li = [11,22] #判断某个对象是否是某个类创建的. r = isinstance(li, list) print(r) 结果: C:\Python ...
- tar打包tar.gz文件
命令格式: tar zcvf dir.tar.gz ./dir 压缩后的文件解压出来会是dir这个文件夹
- 激光SLAM Vs 视觉SLAM
博客转载自:https://www.leiphone.com/news/201707/ETupJVkOYdNkuLpz.html 雷锋网(公众号:雷锋网)按:本文作者SLAMTEC(思岚科技公号sla ...
- Java-马士兵设计模式学习笔记-工厂模式-简单工厂
一.概述 1.目标:要控制任意类型交通工具的生产模式 2.目标有两层意思(1)任意类型 (2)生产模式,所以对应的,要这两个层面上抽象(Movable,VehicleFactory),利用接口,实现多 ...
- js实现上传单个文件
js上传文件:js 上传单个文件(任意大小) 疯狂代码 http://www.CrazyCoder.cn/ :http:/www.CrazyCoder.cn/Javascript/Article832 ...
- CF 1029E Tree with Small Distances
昨晚随便玩玩搞个div3结果浪翻了…… 强烈谴责D题hack数据卡常 考虑到本题中所要求的最短距离不会大于2,所以我们可以把所有结点到$1$的距离通过对$3$取模分类,考虑到直接自顶向下贪心不满足局部 ...
- Luogu 4323 [JSOI2016]独特的树叶
新技能get 树哈希,考虑到两棵树相同的条件,把每一个结点的哈希值和树的siz写进哈希值里去. 做出A树每一个结点为根时的树的哈希值丢进set中,然后暴力枚举B树中度数为1的点,求出删掉这个点之后的哈 ...
- 安装Maven及Eclipse中配置Maven
下载maven版本: 1.进入官网:http://maven.apache.org/download.cgi ,下载编译后的maven版本:如图下: 2.创建一个目录,把下载的maven压缩包,进 ...