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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#122880 | #1808. Efficient Partitioning | Scarlett_boy | WA | 1ms | 3472kb | C++17 | 3.3kb | 2023-07-11 13:27:57 | 2023-07-11 13:27:58 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
int n;
const int mod = 1e9 + 7;
const ll Eps = 1e18, L = 1, R = 2e18;
#define int long long
struct Tree {
#define LOG 65
int root, cnt;
ll Mi, Ma;
vector<int> ls, rs;
vector<ll> MAX1, MAX2;
Tree(int M = 0, ll Min = 1, ll Max = 2e18) : ls(M * LOG), rs(M * LOG), MAX1(M * LOG), MAX2(M * LOG) {
root = 0, cnt = 0;
Mi = Min, Ma = Max;
}
inline void pushup(int p) {
MAX1[p] = max(MAX1[ls[p]], MAX1[rs[p]]);
MAX2[p] = max(MAX2[ls[p]], MAX2[rs[p]]);
}
inline void New(int &p) {
p = ++cnt;
MAX1[p] = MAX2[p] = -Eps;
}
inline void update(int &p, ll l, ll r, ll x, ll w1, ll w2) {
if (!p) New(p);
if (l == r) {
MAX1[p] = max(MAX1[p], w1);
MAX2[p] = max(MAX2[p], w2);
return;
}
int mid = l + r >> 1;
if (x <= mid) update(ls[p], l, mid, x, w1, w2);
else update(rs[p], mid + 1, r, x, w1, w2);
pushup(p);
}
inline void update(int x, ll w1, ll w2) { update(root, Mi, Ma, x, w1, w2); }
inline ll query1(int p, ll l, ll r, ll x, ll y) {
if (!p) return -Eps;
if (x <= l && r <= y) return MAX1[p];
ll ans = -Eps;
int mid = l + r >> 1;
if (x <= mid) ans = max(ans, query1(ls[p], l, mid, x, y));
if (y > mid) ans = max(ans, query1(rs[p], mid + 1, r, x, y));
return ans;
}
inline ll query1(ll l, ll r) { return query1(root, Mi, Ma, l, r); }
inline ll query2(int p, ll l, ll r, ll x, ll y) {
if (!p) return -Eps;
if (x <= l && r <= y) return MAX2[p];
ll ans = -Eps;
int mid = l + r >> 1;
if (x <= mid) ans = max(ans, query2(ls[p], l, mid, x, y));
if (y > mid) ans = max(ans, query2(rs[p], mid + 1, r, x, y));
return ans;
}
inline ll query2(ll l, ll r) { return query2(root, Mi, Ma, l, r); }
};
void solve() {
cin >> n;
vector<ll> a(n + 5), b(n + 5), c(n + 5), sum(n + 5);
vector<ll> dp(n + 5, -Eps);
Tree T(n);
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = 0; i < n; i++) cin >> b[i];
for (int i = 0; i < n; i++) cin >> c[i];
sum[0] = a[0];
for (int i = 1; i < n; i++) sum[i] = a[i] + sum[i - 1];
dp[0] = a[0] + b[0] + c[0];
T.update(dp[0] - b[1] + sum[0] + Eps, dp[0], b[1] - sum[0]);
for (int i = 1; i < n; i++) {
// for (int j = 0; j < i; j++) {
// dp[i] = max(dp[i], min(dp[j], b[j + 1] + c[i] + query(j + 1, i)));
// }
ll x = c[i] + sum[i] + Eps;
dp[i] = max(T.query1(L, x), T.query2(x, R) + c[i] + sum[i]);
// cout << T.query1(L, x) << " " << T.query2(x, R) + c[i] + sum[i] << " " << T.query2(x, R) << "\n";
T.update(dp[i] - b[i + 1] + sum[i] + Eps, dp[i], b[i + 1] - sum[i]);
}
cout << dp[n - 1] << '\n';
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int _ = 1;
// cin >> _;
for (int o = 1; o <= _; o++) {
// cout << "Case #<<" << o << ": ";
solve();
}
return 0;
}
/*
1
10 4
AABABBABAB
10 3 4 10 7 1 8 10 7 6
*/
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 0ms
memory: 3420kb
input:
2 1 -1 -1 4 1 -2
output:
1
result:
ok answer is '1'
Test #2:
score: 0
Accepted
time: 1ms
memory: 3472kb
input:
1 1000000000 1000000000 1000000000
output:
3000000000
result:
ok answer is '3000000000'
Test #3:
score: -100
Wrong Answer
time: 1ms
memory: 3472kb
input:
11 -323225375 -897098227 -795978453 501188072 409939326 -362890219 969123048 962633819 252457646 694824070 -406990840 -696821643 -663750226 -570551722 670541392 172964990 399404695 -305728788 -157617655 -801518744 -328729631 -160335217 -465411342 -660775657 515997870 -34787742 628368976 84800619 -72...
output:
787996627
result:
wrong answer expected '91174984', found '787996627'