CF1699D. Almost Triple Deletions (DP)

https://codeforces.com/contest/1699/problem/D
题意： 每次操作删除相邻且不同的两个数，问最后留下的最长等元素数组的长度。数组长度 1 <= n <= 1000.
思路:

• 假设最终数组留下的数字是x, 那么所有x的位置pos之间和pos的到边界的数都应该删完。
• 一段区间[l, r]的数要删完，区间中出现最多的数字的出现次数mx * 2 <= 区间长度len, 并且len必须是偶数
• 枚举最后的数字x，看x之间的数能否删完。这样想比较麻烦，思考可以发现这个问题有阶段性，不妨借助dp方程转移
• dp[i]表示以第i个位置上的数做最终序列的结尾的最长长度

#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false) ,cin.tie(0), cout.tie(0);
//#pragma GCC optimize(3,"Ofast","inline")
#define ll long long
#define PII pair<int, char>
//#define int long long
const int N = 5e3 + 5;
const int M = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const ll LNF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
int f[N][N], a[N], dp[N];
bool check( int l, int r ) {
  if( r - l < 0 ) return true;
  if( f[l][r] <= (r - l + 1) / 2 && (r - l + 1) % 2 == 0 ) return true;
  return false;
}
void solve() {
  int n; cin >> n;
  for ( int i = 1; i <= n; ++ i ) cin >> a[i];
  for ( int i = 1; i <= n; ++ i ) {
    vector<int> c(n + 1, 0);
    for ( int j = i; j <= n; ++ j ) {
      f[i][j] = max( f[i][j - 1], ++ c[a[j]]);
    }
  }
  for ( int i = 1; i <= n; ++ i ) {
    if( check( 1, i - 1 ) ) {
      dp[i] = 1;
    }
  }
  for ( int i = 1; i <= n; ++ i) {
    for ( int j = 1; j <= i - 1; ++ j ) {
      if( a[i] == a[j] && check( j + 1, i - 1) ) dp[i] = max(dp[i], dp[j] + 1);
    }
  }
  int ans = 0;
  for ( int i = 1; i <= n; ++ i) {
    if( check( i + 1, n) ) ans = max( ans , dp[i]);
  }
  cout << ans << '\n';
  for ( int i = 1; i <= n; ++ i ) {
    dp[i] = -0x3f3f3f3f;
    for ( int j = i; j <= n; ++ j ) {
      f[i][j] = 0;
    }
  }
}
int main () {
  memset(dp, -0x3f, sizeof dp);
  int t; cin >> t;
  while ( t -- ) solve();
  return 0;
}

标签：Deletions,const,Almost,long,int,dp,数组,DP,define
来源： https://www.cnblogs.com/muscletear/p/16445633.html