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

巴特西

hdu6438 Buy and Resell 买卖物品 ccpc网络赛贪心

Buy and Resell

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巴特西

hdu6438 Buy and Resell 买卖物品 ccpc网络赛 贪心

Buy and Resell

最新文章

热门文章

hdu6438 Buy and Resell 买卖物品 ccpc网络赛贪心