hdu-6438-貪心
阿新 • • 發佈：2018-08-26
inf cat show col .cn 獲得 day 隊列 esc
1. spend ai dollars to buy a 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.
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.
Buy and Resell
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1160 Accepted Submission(s): 369
1. spend ai dollars to buy a 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 2018中國大學生程序設計競賽 - 網絡選拔賽 用一個優先隊列維護當前可購買的最小值有哪些，對每一天考慮如果當前這天賣出可獲得的最大價值，由於在當前天賣出不一定是最優的，在賣出後在將a的值入隊然後標記一下表示這個a是由另一個價值x賣掉之後得到的，和普通的a略有不同，這樣在後面出現a是最小值的 時候次數就可以不變了。(x->a->y <==> x->y)，相同的數可能出現多次所以標記可以累計，一開始只是用了0/1瘋狂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-貪心