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  1. #include <bits/stdc++.h>
  2.  
  3. using namespace std;
  4.  
  5. #define ll long long
  6. #define FOR(i , a , b) for(int i = a ; i <= b; i++)
  7. #define FAST ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  8. #define maxn 505
  9.  
  10. int a[maxn] , b[maxn] , dp[maxn][maxn][2] , pre[maxn][maxn][2];
  11.  
  12. int main(){
  13.     FAST;
  14.     int n , m;
  15.     cin >> n >> m;
  16.     FOR(i , 1 , n) cin >> a[i];
  17.     FOR(i  ,1 , m) cin >> b[i];
  18.  
  19.     // goi dp[i][j][k] la do dai day bap benh chung dai nhat khi xet a[1.. i] va b[1..j] 
  20.     // ai > bj -> k = 1
  21.     // ai < bj -> k = 0
  22.     FOR(i , 1 , n){
  23.         FOR(j , 1 , m){
  24.             if(a[i] != b[j]) continue;
  25.             int best_greater = 1;
  26.             int best_less = 1;
  27.  
  28.             // dp[i][j][0] = max(dp[i][j][0] , dp[i'][j'][1] + 1)
  29.             // dp[i][j][1] = max(dp[i][j][1] , dp[i'][j'][0] + 1)
  30.             // ta thay rang ta se xet i' < i va j' < j -> ta co dinh chieu i va opt j
  31.             // -> opt j bang prefix sum
  32.             // pre[i][j][0] , pre[i][j][1]
  33.             // pre[i][j][0] = max(pre[i][j - 1][0] , dp[i][j][0])
  34.  
  35.             FOR(id , 1 , i - 1){
  36.                 if(a[id] < a[i]){
  37.                     int cand = pre[id][j - 1][0];
  38.                     if(cand > 0) best_greater = max(best_greater , cand + 1);
  39.                 }
  40.  
  41.                 else if(a[id] > a[i]){
  42.                     int cand = pre[id][j - 1][1];
  43.                     if(cand > 0) best_less = max(best_less , cand + 1);
  44.                 }
  45.             }
  46.             dp[i][j][0] = best_less;
  47.             dp[i][j][1] = best_greater;
  48.         }
  49.  
  50.         pre[i][0][0] = 0;
  51.         pre[i][0][1] = 0;
  52.         // sau khi build xong tung j cho hang i
  53.         FOR(j , 1 , m){
  54.             pre[i][j][0] = max(pre[i][j - 1][0] , dp[i][j][0]);
  55.             pre[i][j][1] = max(pre[i][j - 1][1] , dp[i][j][1]);
  56.         }
  57.     }
  58.  
  59.     int ans = 0;
  60.     FOR(i , 1 , n){
  61.         FOR(j , 1 , m){
  62.             ans = max({ans , dp[i][j][0] , dp[i][j][1]});
  63.         }
  64.     }
  65.  
  66.     cout << ans;
  67. }
Success #stdin #stdout 0.01s 5284KB
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https://ideone.com/S9bRVL
language:
C++14 (gcc 8.3)
created:
9 hours ago
visibility:
public

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