牛客网暑期ACM多校训练营（第二场） D: money

链接：https://www.nowcoder.com/acm/contest/140/D
来源：牛客网
 

White Cloud has built n stores numbered from 1 to n.
White Rabbit wants to visit these stores in the order from 1 to n.
The store numbered i has a price a[i] representing that White Rabbit can spend a[i] dollars to buy a product or sell a product to get a[i] dollars when it is in the i-th store.
The product is too heavy so that White Rabbit can only take one product at the same time.
White Rabbit wants to know the maximum profit after visiting all stores.
Also, White Rabbit wants to know the minimum number of transactions while geting the maximum profit.
Notice that White Rabbit has infinite money initially.

输入描述:

The first line contains an integer T(0<T<=5), denoting the number of test cases.
In each test case, there is one integer n(0<n<=100000) in the first line,denoting the number of stores.
For the next line, There are n integers in range [0,2147483648), denoting a[1..n].

输出描述:

For each test case, print a single line containing 2 integers, denoting the maximum profit and the minimum number of transactions.

示例1

输入

1
5
9 10 7 6 8

输出

3 4

题意：

给你一个[1,n]的商店顺序，你只能从1-n，你可以在每个商店选择买东西或者卖东西，买卖的价钱就是商店的权值，求最多大利润和最小的操作数

思路：

就是一个简单的dp，每次记录前面的sell最优和buy最优

	9	10	7	6	8
buy	-9	-10	-6	-5	-7
sell	0	1	-2	0	3
bmax	-9	-9	-6	-5	-5
smax	0	1	1	1	3

如图所示，对于每一件商品，都维护smax和bmax的值，下一次的buy值和sell值有smax和bmax得到最后得到的smax就只最大利益

而对于操作数可以类似

#include <iostream>
#include <string.h>
using namespace std;
#define ll long long
ll dp[100005][2];
int main() {
    ios::sync_with_stdio(false);
    int T;
    cin >> T;
    for (int i = 1; i <= T; i ++) {
        int n;
        cin >> n;
        int x;
        cin >> x;
        memset(dp, 0, sizeof(dp));
        ll bmax, smax;
        bmax = (-1) * x;
        smax = 0;
        int ds = 0, db = 1;
        for (int j = 2, x; j <= n; j ++) {
            cin >> x;
            dp[j][0] = smax - x;//buy
            dp[j][1] = bmax + x;//sell
            if (smax < dp[j][1]) {
                ds = db + 1;//更新sell的操作数
                smax = dp[j][1];//更新smax
            }
            if (bmax < dp[j][0]) {
                db = ds + 1;//更新buy的操作数
                bmax = dp[j][0];//更新bmax
            }
        }
        cout << smax << " " << ds <<endl;
    }
    return 0;
}