DMY 周作业 47 简要题解

G

数据结构优化 DP 板。暑假的时候做过，直接离散化 + BIT 就行了。比较无聊就不说了。

#include <bits/stdc++.h>
#define fi first
#define se second
#define eb(x) emplace_back(x)
#define pb(x) push_back(x)
#define lc(x) (tr[x].ls)
#define rc(x) (tr[x].rs)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
using pi = pair<int, int>;
const int N = 200005;
const ll inf = 0x3f3f3f3f3f3f3f3f;
int n;
ll a[N], dp[N], lsh[N], cnt;
int lowbit(int x)
{return (x & (-x));
}
struct BIT{ll tr[N];void init(){memset(tr, -0x3f, sizeof(tr));}void update(int p, ll v){while(p <= cnt){tr[p] = max(tr[p], v);p += lowbit(p);}}ll query(int p){ll res = -inf;while(p){res = max(res, tr[p]);p -= lowbit(p);}return res;}
}tr1;
int getid(ll x)
{return (lower_bound(lsh + 1, lsh + cnt + 1, x) - lsh);
}
int main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin >> n;for(int i = 1; i <= n; i++){cin >> a[i];a[i] += a[i - 1];lsh[++cnt] = a[i];}lsh[++cnt] = 0;sort(lsh + 1, lsh + cnt + 1);cnt = unique(lsh + 1, lsh + cnt + 1) - lsh - 1;tr1.init();tr1.update(getid(0), 0);ll premx = 0;for(int i = 1; i <= n; i++){dp[i] = tr1.query(getid(a[i])) + i;premx = max(premx, dp[i]);tr1.update(getid(a[i]), premx - i);}cout << premx;return 0;
}