DP Codeforces Round #260 (Div. 1) A. Boredom

题目传送门

 1 /*
 2     题意：选择a[k]然后a[k]-1和a[k]+1的全部删除，得到点数a[k]，问最大点数
 3     DP：状态转移方程：dp[i] = max (dp[i-1], dp[i-2] + (ll) i * cnt[i]);
 4         只和x-1，x-2有关，和顺序无关，x-1不取，x-2取那么累加相同的值，ans = dp[mx]
 5 */
 6 #include <cstdio>
 7 #include <algorithm>
 8 #include <cstring>
 9 #include <iostream>
10 #include <cmath>
11 using namespace std;
12
13 typedef long long ll;
14 const int MAXN = 1e5 + 10;
15 const int INF = 0x3f3f3f3f;
16 ll dp[MAXN];
17 int cnt[MAXN];
18
19 int main(void)        //Codeforces Round #260 (Div. 1) A. Boredom
20 {
21     int n;
22     while (scanf ("%d", &n) == 1)
23     {
24         memset (dp, 0, sizeof (dp));
25         memset (cnt, 0, sizeof (cnt));
26
27         int x, mx = 0;
28         for (int i=1; i<=n; ++i)    {scanf ("%d", &x);    cnt[x]++;    if (mx < x) mx = x;}
29
30         dp[1] = cnt[1];
31         for (int i=2; i<=mx; ++i)
32         {
33             dp[i] = max (dp[i-1], dp[i-2] + (ll) i * cnt[i]);
34         }
35
36         printf ("%I64d\n", dp[mx]);
37     }
38
39     return 0;
40 }
41
42
43
44 /*
45 2
46 1 2
47 3
48 1 2 3
49 9
50 1 2 1 3 2 2 2 2 3
51 */