标签：positive das nbsp div cin mini dash font 设计
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1160 Accepted Submission(s): 369
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
用一个优先队列维护当前可购买的最小值有哪些，对每一天考虑如果当前这天卖出可获得的最大价值，由于在当前天卖出不一定是最优的，在卖出后在将a的值入队然后标记一下表示这个a是由另一个价值x卖掉之后得到的，和普通的a略有不同，这样在后面出现a是最小值的
时候次数就可以不变了。(x->a->y <==> x->y)，相同的数可能出现多次所以标记可以累计，一开始只是用了0/1疯狂WA。
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define LL long long
4 map<int,int>M;
5 int main(){
6 int t,n,m,i,j,k,a;
7 cin>>t;
8 while(t--){
9 scanf("%d",&n);
10 priority_queue<int,vector<int>,greater<int> >q;
11 M.clear();
12 LL s1=0,s2=0;
13 for(i=1;i<=n;++i){
14 scanf("%d",&a);
15 if(q.empty() || q.top()>=a){
16 q.push(a);
17 }
18 else{
19 s1+=a-q.top();
20 if(M[q.top()]==0){
21 s2++;
22 }
23 else{
24 M[q.top()]--;
25 }
26 q.pop();
27 q.push(a);
28 q.push(a);
29 M[a]++;
30 }
31
32 }
33 //cout<<s1<<endl;
34 printf("%lld %lld\n",s1,s2*2);
35 }
36 return 0;
37 }
hdu-6438-贪心
标签：positive das nbsp div cin mini dash font 设计
原文地址：https://www.cnblogs.com/zzqc/p/9537656.html
