Two Melodies CodeForces - 813D (DP,技巧)

https://codeforces.com/problemset/problem/813/D

dp[i][j] = 一条链以i结尾, 另一条链以j结尾的最大值

关键要保证转移时两条链不能相交

#include <iostream>
#include <cstdio>
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;
const int N = 5e3+, M = 1e7+;
int n;
int a[N], dp[N][N], x[M], y[M];
 
int main() {
    cin>>n;
    REP(i,,n) cin>>a[i];
    int ans = ;
    REP(i,,n) {
        REP(j,,) x[j]=;
        REP(j,,n) y[a[j]]=;
        REP(j,,i) {
            x[a[j]%]=max(x[a[j]%],dp[i][j]);
            y[a[j]]=max(y[a[j]],dp[i][j]);
        }
        REP(j,i+,n) {
            dp[i][j]=max(dp[i][],max(y[a[j]+],y[a[j]-]));
            dp[i][j]=max(dp[i][j],x[a[j]%]);
            dp[j][i]=++dp[i][j];
            x[a[j]%]=max(x[a[j]%],dp[i][j]);
            y[a[j]]=max(y[a[j]],dp[i][j]);
            ans = max(ans, dp[i][j]);
        }
    }
    printf("%d\n", ans);
}