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

题目传送门

 /*
     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
     DP：状态转移方程：dp[i] = max (dp[i-1], dp[i-2] + (ll) i * cnt[i]);
         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]
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <iostream>
 #include <cmath>
 using namespace std;

 typedef long long ll;
 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 ll dp[MAXN];
 int cnt[MAXN];

 int main(void)        //Codeforces Round #260 (Div. 1) A. Boredom
 {
     int n;
     while (scanf ("%d", &n) == )
     {
         memset (dp, , sizeof (dp));
         memset (cnt, , sizeof (cnt));

         int x, mx = ;
         for (int i=; i<=n; ++i)    {scanf ("%d", &x);    cnt[x]++;    if (mx < x) mx = x;}

         dp[] = cnt[];
         for (int i=; i<=mx; ++i)
         {
             dp[i] = max (dp[i-], dp[i-] + (ll) i * cnt[i]);
         }

         printf ("%I64d\n", dp[mx]);
     }

     return ;
 }

 /*
 2
 1 2
 3
 1 2 3
 9
 1 2 1 3 2 2 2 2 3
 */