CF456C Boredom （DP）

Boredom

CF#260 div2 C. Boredom

Codeforces Round #260

C. Boredom time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it. Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak points to the player. Alex is a perfectionist, so he decided to get as many points as possible. Help him. Input The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex's sequence. The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105). Output Print a single integer — the maximum number of points that Alex can earn. Sample test(s) Input 2 1 2 Output 2 Input 3 1 2 3 Output 4 Input 9 1 2 1 3 2 2 2 2 3 Output 10 Note Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.

题意：给出一堆数，一次操作是取某个数，然后等于(它+1)或(它-1)的数全删掉，得到(它)分，求最高得分。

有两种dp方法，一种排序后dp[i][j]表示取到第i个得到的最大分，j表示取或不取；

另一种是我用的，dp[i]表示在1~i的数中取，必取i，得到的最高分，从2扫到100000。

            dp[i]=dp[i-];
            if(a[i]>)dp[i]=max(dp[i],dp[i-]+1LL*i*a[i]);

注意10^5 * 10^5= 10^10会爆long long，我就是这样掉分飞起的。

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll long long
 #define usint unsigned int
 #define mz(array) memset(array, 0, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(i=0;i<(n);i++)
 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) printf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("1biao.out","w",stdout)

 const int maxn=;
 int a[maxn];
 ll dp[maxn];

 int main(){
     ll ma;
     int n,i,x;
     while(scanf("%d",&n)!=EOF){
         mz(a);
         for(i=;i<n;i++){
             scanf("%d",&x);
             a[x]++;
         }
         dp[]=;
         dp[]=a[];
         ma=dp[];
         for(i=;i<=;i++)
         {
             dp[i]=dp[i-];
             if(a[i]>)dp[i]=max(dp[i],dp[i-]+1LL*i*a[i]);
             if(dp[i]>ma)ma=dp[i];
         }
         printf("%I64d\n",ma);
     }
     return ;
 }