CodeTON Round 5 (Div. 1 + Div. 2, Rated, Prizes!)C

合集 - acm(93)

1.AtCoder Beginner Contest 293(C,D ,E,F)2023-03-172.Educational Codeforces Round 115 (Rated for Div. 2)（D,E)2023-03-183.AtCoder Beginner Contest 294(E,F,G)2023-04-034.AtCoder Beginner Contest 2462023-03-295.中国石油大学（北京）第三届“骏码杯”程序设计竞赛（同步赛）(D,E,F)2023-03-256.Codeforces Global Round 16(D,E,F)2023-03-257.AtCoder Beginner Contest 209（D，E）2023-05-088.Monoxer Programming Contest 2022（AtCoder Beginner Contest 238）（E，F）2023-05-079.AtCoder Beginner Contest 285（B,D,E,F)2023-05-0610.AtCoder Beginner Contest 242（D，E)2023-05-0311.AtCoder Beginner Contest 223（D,E,F)2023-04-1512.AtCoder Beginner Contest 207（D,E)2023-04-1113.AtCoder Beginner Contest 247（E,F)2023-04-1014.AtCoder Beginner Contest 226(E,F,G)2023-04-0515.AtCoder Beginner Contest 229(F,G)2023-06-2316.AtCoder Beginner Contest 273(E)2023-06-1317.AtCoder Beginner Contest 286(G)2023-06-0218.AtCoder Beginner Contest 287(C,D,E,F)2023-06-0119.AtCoder Beginner Contest 288(D,E,F)2023-05-3120.AtCoder Beginner Contest 289(E,F)2023-05-3021.AtCoder Beginner Contest 290(D,E)2023-05-2922.AtCoder Beginner Contest 292(E,F,G)2023-05-2823.AtCoder Beginner Contest 298(D,F)2023-05-2724.AtCoder Beginner Contest 299(E,F)2023-05-2725.AtCoder Beginner Contest 300(E,F)2023-05-2526.AtCoder Beginner Contest 302（E，F，G）2023-05-2427.AtCoder Beginner Contest 253(E,F)2023-05-1728.AtCoder Beginner Contest 245(D，E，F)2023-05-1529.AtCoder Beginner Contest 248(D，E，F) 2023-05-1430.AtCoder Beginner Contest 206（Sponsored by Panasonic）（E，F）2023-05-0931.Codeforces Round 875 (Div. 2)(D)2023-07-0832.AtCoder Beginner Contest 178(E,F)2023-07-0833.AtCoder Beginner Contest 307（E,F,G)2023-07-01

34.CodeTON Round 5 (Div. 1 + Div. 2, Rated, Prizes!)C2023-06-30

35.Educational Codeforces Round 151 (Rated for Div. 2)(C,D)2023-06-3036.AtCoder Beginner Contest 212(E,F)2023-06-2437.牛客小白月赛512022-12-0938.2022年浙大城市学院新生程序设计竞赛（同步赛）（补题）2022-12-1139.Codeforces Round #770 (Div. 2)B，C2022-12-2440.Codeforces Global Round 24(B,C)2022-12-2341.Codeforces Round #836 (Div. 2）C2022-12-2142.Codeforces Round #840 (Div. 2) C2022-12-2043.Good Bye 2022: 2023 is NEAR C2022-12-3144.Codeforces Round #765 (Div. 2)A，B，C2022-12-3145.Codeforces Round #766 (Div. 2)C，D2022-12-2946.Codeforces Round #841 (Div. 2) and Divide by Zero 20222022-12-2847.Codeforces Round #767 (Div. 2)C ,D 2022-12-2748.Codeforces Round #768 (Div. 2)C ,D2022-12-2649. Codeforces Round #769 (Div. 2) B，C2022-12-2550.Educational Codeforces Round 122 (Rated for Div. 2)，C，D2022-12-2551.Educational Codeforces Round 119 (Rated for Div. 2)2023-01-1452.AtCoder Beginner Contest 2582023-01-1353.Codeforces Round #763 (Div. 2)C2023-01-1254.Codeforces Round #843 (Div. 2)（B，C，D，E）2023-01-1255.Educational Codeforces Round 141 (Rated for Div. 2)（B，C，D）2023-01-1056.AtCoder Beginner Contest 275（B，C，D，E，F）2023-01-0957.Codeforces Round #842 (Div. 2)（B，D，E）2023-01-0958.AtCoder Beginner Contest 284（D，E，F）2023-01-0859.The 14th Jilin Provincial Collegiate Programming Contest（补题）2023-01-0760.牛客小白月赛65（C，D，E，F）2023-01-0761.AtCoder Beginner Contest 281（D，E，F）2023-01-0562.Good Bye 2021: 2022 is NEAR D2023-01-0563.Hello 20232023-01-0464.The 15th Jilin Provincial Collegiate Programming Contest(补题)2023-01-0365.Codeforces Round #781 (Div. 2)C 2023-01-0266.Hello 2022（B，D）2023-01-0167.AtCoder Beginner Contest 272（D,E)2023-03-1068.Codeforces Round 751 (Div. 2)（D）2023-03-0869.Codeforces Round 856 (Div. 2)(C,D)2023-03-0870.Codeforces Round 752 (Div. 2)（C，D，E）2023-03-0871.Codeforces Round 855 (Div. 3)（E，F）2023-03-0572.AtCoder Regular Contest 131（A,B,C）2023-03-0573.Educational Codeforces Round 144 (Rated for Div. 2)（A，B，C，D）2023-03-0574.Codeforces Round 853 (Div. 2)（C,D)2023-03-0375.牛客练习赛109（C,D)2023-03-0376.AtCoder Beginner Contest 291（Sponsored by TOYOTA SYSTEMS）（D，E，F）2023-03-0177.Educational Codeforces Round 143 (Rated for Div. 2)（A，C,D)2023-02-2878.Codeforces Round #852 (Div. 2)（C,D)2023-02-1379.Educational Codeforces Round 118 (Rated for Div. 2)（D，E）2023-02-0980.AtCoder Beginner Contest 236（D，E，F）2023-02-0981.Codeforces Round #850 (Div. 2, based on VK Cup 2022 - Final Round)(B,D)2023-02-0882.Codeforces Round #848 (Div. 2)（B,C,D）2023-02-0783.TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) (B,C,D) 2023-02-0784.Codeforces Round #846 (Div. 2)（B，E） 2023-02-0785.Educational Codeforces Round 142 (Rated for Div. 2)（C，D）2023-02-0486.2023牛客寒假算法基础集训营62023-02-0487.Codeforces Round #845 (Div. 2) and ByteRace 2023（A，B，C）2023-02-0288.2023牛客寒假算法基础集训营5 2023-02-0289.2023牛客寒假算法基础集训营3 2023-01-2190.2023牛客寒假算法基础集训营22023-01-1991.2023牛客寒假算法基础集训营12023-01-1892.Educational Codeforces Round 120 (Rated for Div. 2) C，D2023-01-1593.AtCoder Beginner Contest 254（C，D，E，F) 2023-01-15

收起

CodeTON Round 5 (Div. 1 + Div. 2, Rated, Prizes!)C

C(dp)

C

题目给出一个数组，我们可以在这一个数组里面找出$a_i$和$a_j$其中$i$和$j$不相等，并保证$a_i=a_j$,这样我们就可以把$i-j$这一段区间的所有数字都删除，问我们怎样删除才可以使删除的数字数量最多

很明显，这是一个$dp$，但是如果对于每一对相同数字，根据这个得到不同的线段，然后一个一个的求，会$TLE$，所以我们需要一个特殊的操作让我们只需要那个最优的情况（求$i$之前的所有的$max(d{p}_j)$会超时，但是如果预处理出$m{x}_i=max(d{p}_i)$,但是我们不知道这个$m{x}_{i-1}$的开始在哪一个位置，无法求出这一段的贡献

我们可以设置$d{p}_i$前$i$个位置的最大删除数量为多少

然后我们这里我预处理的$m{x}_i=max(d{p}_i-i)$,这样就相当于把$d{p}_i$在$i$的位置看成往左挪动到位置$0$，这样，后面如果我们用到它，我们只要知道右端点即可，即$d{p}_{i+1}=m{x}_{a[i]}+i+1$,代表着$0$到当前位置$i$的距离，也可以理解为$d{p}_{i+1}=d{p}_{i-1}+i-(i-1)+1$(如果这一个mx_a_i是从$d{p}_{i-1}$那里得来的)

这样就很好写了

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include<cmath>
#include <unordered_map>
#include <array>
#include <cstring>
#include <bitset>
#include <numeric>
using namespace std;
#define int long long 
#define LL long long
#define ios  ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define inf 1e18
#define INF 1e18
#define eps 1e-6
#define mem(a,b) memset((a),(b),sizeof(a))
const int maxn=2e5+10;
const int mod=998244353;
int t;
int n,a[maxn];
void solve()
{
  cin>>n;
  for (int i=1;i<=n;i++)
  {
    cin>>a[i];
  }
  vector<int>dp(n+2,0),mx(n+2,-inf);
  for(int i=1;i<=n;i++)
  {
    dp[i+1]=max(dp[i],mx[a[i]]+i+1);
    mx[a[i]]=max(mx[a[i]],dp[i]-i);//mx记录的是在某一个数，并以其中一个位置作为起点的最大值，这样，后面如果要用到这一个数，我们可以直接把mx+pos+1,因为我们之前已经减去了本身的距离
  }
  int ans=dp[n+1];
  cout<<ans<<"\n";
  return ;
}
signed main ()
{
  //ios;
  cin>>t;
  while (t--)
  {
    solve();
  }
  system ("pause");
  return 0;
}