百练4121 股票买卖

Posted 2020-08-30 王宜鸣

思路：

dp。

实现：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 const int MAXN = 100005;
 7 const int INF = 0x3f3f3f3f;
 8 int dp[5], n, T, p[MAXN];
 9 
10 int solve()
11 {
12     for (int i = 0; i < 4; i++)
13     {
14         dp[i] = -INF;
15     }
16     dp[4] = 0;
17     for (int i = n - 1; i >= 0; i--)
18     {
19         dp[0] = max(dp[4], max(dp[3] - p[i], max(dp[2], max(dp[0], dp[1] - p[i]))));
20         dp[1] = max(dp[4] + p[i], max(dp[3], max(dp[1], dp[2] + p[i])));
21         dp[2] = max(dp[4], max(dp[2], dp[3] - p[i]));
22         dp[3] = max(dp[3], dp[4] + p[i]);
23         dp[4] = dp[4];
24     }
25     return dp[0];
26 }
27 int main()
28 {
29     scanf("%d", &T);
30     while (T--)
31     {
32         scanf("%d", &n);
33         for (int i = 0; i < n; i++)
34         {
35             scanf("%d", &p[i]);
36         }
37         cout << solve() << endl;
38     }
39     return 0;
40 }