Buy and Resell
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2048 Accepted Submission(s): 745
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
Source
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题意:给出n个城市的物价,每个城市可以买也可以卖,求出在最大利润的条件下最小的交易次数。
题解:用set维护前面的物价,当输入一个x时,若大于set的首元素,则就在x这里卖出set的首元素,同时将x加入set两次,先加入一个交易代价为0的,再加入交易代价为1的,意思是可以反悔,一旦遇到比这个更大的便可以视作做这个x收回,在更大的这里进行交易。
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<set>
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define SI(i) scanf("%lld",&i)
#define PI(i) printf("%lld\n",i)
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int MAX=2e5+5;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
int dir[9][2]={0,1,0,-1,1,0,-1,0, -1,-1,-1,1,1,-1,1,1};
template<class T>bool gmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>bool gmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>void gmod(T &a,T b){a=((a+b)%mod+mod)%mod;}
typedef pair<ll,ll> PII;
int a[MAX];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
multiset<PII>s;
ll ans=0,cnt=0;
for(int i=0;i<n;i++)
{
ll x;
SI(x);
if(!s.empty() && x>s.begin()->first)
{
ans+=x-s.begin()->first;
cnt+=s.begin()->second*2;
s.erase(s.begin());
s.insert(make_pair(x,0));
s.insert(make_pair(x,1));
}
else
{
s.insert(make_pair(x,1));
}
}
printf("%lld %lld\n",ans,cnt);
}
return 0;
}
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