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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#382362 | #7901. Basic Substring Structure | EBeason | TL | 8ms | 86636kb | C++20 | 7.4kb | 2024-04-08 13:07:29 | 2024-04-08 13:07:29 |
Judging History
answer
#pragma GCC optimize(1, 2, 3, "Ofast", "inline")
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned ll
// #define int ll
#define ls p << 1
#define rs p << 1 | 1
#define lowbit(x) ((x) & (-x))
#define endl '\n'
#define ld long double
#define MULTI_CASES
const int MaxN = 1e6 + 100;
const int INF = 1e9;
int T, N, M, K;
int ans[MaxN];
vector<int> a;
struct SA {
int sa[MaxN], rk[MaxN], oldrk[MaxN << 1], id[MaxN], key1[MaxN], cnt[MaxN], height[MaxN];
int st[MaxN][25], logn[MaxN];
int n;
vector<int> s;
void init(int _n) {
n = _n;
for (int i = 0; i <= n + 1; i++) {
sa[i] = 0;
rk[i] = 0;
oldrk[i] = 0;
cnt[i] = 0;
key1[i] = 0;
id[i] = 0;
height[i] = 0;
}
getsa();
getheight();
}
bool cmp(int x, int y, int w) {
return oldrk[x] == oldrk[y] && oldrk[x + w] == oldrk[y + w];
}
inline void getsa() {
int i, m = n + 1, p, w;
// sa[i]表示后缀排序第i小的编号
// rk[i]表示后缀i的排名
for (i = 1; i <= n; ++i) ++cnt[rk[i] = s[i]];
for (i = 1; i <= m; ++i) cnt[i] += cnt[i - 1];
for (i = n; i >= 1; --i) sa[cnt[rk[i]]--] = i;
for (w = 1;; w <<= 1, m = p) { // m=p 就是优化计数排序值域
for (p = 0, i = n; i > n - w; --i) id[++p] = i;
for (i = 1; i <= n; ++i)
if (sa[i] > w) id[++p] = sa[i] - w;
memset(cnt, 0, sizeof(cnt));
for (i = 1; i <= n; ++i) ++cnt[key1[i] = rk[id[i]]];
// 注意这里px[i] != i,因为rk没有更新,是上一轮的排名数组
for (i = 1; i <= m; ++i) cnt[i] += cnt[i - 1];
for (i = n; i >= 1; --i) sa[cnt[key1[i]]--] = id[i];
memcpy(oldrk + 1, rk + 1, n * sizeof(int));
for (p = 0, i = 1; i <= n; ++i) rk[sa[i]] = cmp(sa[i], sa[i - 1], w) ? p : ++p;
if (p == n) break;
}
}
inline void getheight(){
// height[i]=lcp(sa[i],sa[i-1])
int i, k;
for (i = 1, k = 0; i <= n; ++i) {
if (rk[i] == 0) continue;
if (k) --k;
while (i + k <= n && s[i + k] == s[sa[rk[i] - 1] + k]) ++k;
height[rk[i]] = k;
}
}
inline void pre() {
logn[0] = -1;
int LogN = log2(n) + 2;
for (int i = 1; i <= n; i++) logn[i] = logn[i / 2] + 1;
for (int i = 1; i <= n; i++) st[i][0] = height[i];
for (int j = 1; j <= LogN; j++) {
int pj = 1 << (j - 1);
for (int i = 1; i <= n; i++) {
if (i + pj <= n)
st[i][j] = min(st[i][j - 1], st[i + pj][j - 1]);
else
st[i][j] = st[i][j - 1];
}
}
}
inline int lcp(int l, int r) {
if (l == r) return n - l + 1;
l = rk[l];
r = rk[r];
if (l > r) swap(l, r);
l++;
int lp = r - l + 1;
int n = 1 << logn[lp];
return min(st[l][logn[lp]], st[r - n + 1][logn[lp]]);
}
} SA;
struct TT {
int a[MaxN];
struct Point {
ll sum, lan;
} tree[MaxN << 2], ts;
void js(int p, int x, int y, int z) {
tree[p].sum += (y - x + 1) * z;
tree[p].lan += z;
}
void pushdown(int p,int l, int r) {
if (tree[p].lan) {
int mid = (l + r) >> 1;
js(ls, l, mid, tree[p].lan);
js(rs, mid + 1, r, tree[p].lan);
}
}
inline Point pushup(Point L, Point R) {
Point now = Point();
now.sum = L.sum + R.sum;
return now;
}
void build(int p, int l, int r) {
if (l == r) {
tree[p].sum = a[l];
tree[p].lan = 0;
return;
}
pushdown(p, l, r);
int mid = (l + r) >> 1;
build(ls, l, mid);
build(rs, mid + 1, r);
tree[p] = pushup(tree[ls], tree[rs]);
}
void change(int p, int l, int r, int x, int y, int z) {
if (x > y || l > N) return;
if (l >= x && r <= y) {
js(p, l, r, z);
return;
}
pushdown(p, l, r);
int mid = (l + r) >> 1;
if (mid >= x) change(ls, l, mid, x, y, z);
if (mid < y) change(rs, mid + 1, r ,x ,y, z);
tree[p] = pushup(tree[ls], tree[rs]);
}
Point query(int p, int l, int r, int x, int y) {
if (l >= x && r <= y) {
return tree[p];
}
pushdown(p, l, r);
int mid = (l + r) >> 1;
if (mid >= x && mid < y) {
return pushup(query(ls, l, mid, x, y), query(rs, mid + 1, r, x, y));
}
if (mid >= x) return query(ls, l, mid, x, y);
if (mid < y) return query(rs, mid + 1, r, x, y);
return Point();
}
} TT;
ll g[MaxN], b[MaxN];
void updata(int l, int r, int k, int d) {
// cerr << l << ' ';
// cerr << r << ' ';
// cerr << k << ' ';
// cerr << d << endl;
if (l > r) return;
TT.change(1, 1, N, l + 1, r, 1);
TT.change(1, 1, N, l, l, k);
TT.change(1, 1, N, r + 1, r + 1, -k - d * (r - l));
}
struct my_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM =
chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
unordered_map<int ,ll, my_hash> mp[MaxN];
int js(int x, int y) {
// if (a[x] == a[1]) return 0;
int t = SA.lcp(x + 1, y + 1) + 1;
if (x + t >= y) {
int len = y - x;
t = len;
if (x <= N && y + len <= N && a[x] == a[y + len]) {
t += SA.lcp(x + len + 1, y + len + 1) + 1;
}
return t;
} else {
return t;
}
}
inline void Solve() {
cin >> N;
a.assign(N + 1, 0);
for (int i = 1; i <= N; i++) {
cin >> a[i];
mp[i].clear();
}
SA.s = a;
SA.init(N);
SA.pre();
for (int i = 1; i <= N; i++) {
g[i] = SA.lcp(1, i);
// cerr << g[i] << ' ';
}
int sum = 0;
for (int i = 1; i <= N; i++) {
sum += g[i];
TT.a[i] = 0;
}
TT.a[1] = sum;
TT.build(1, 1, N);
for (int i = 1; i <= N; i++) {
int l = 1, r = g[i];
int x = i, y = x + g[i] - 1;
if (i == 1) continue;
if (r < x) {
updata(l, r, -g[i], 1);
updata(x, y, -g[i], 1);
} else if (x <= r) {
updata(l, x - 1, -g[i], 1);
updata(x, y, -g[i], 1);
}
// for (int i = 1; i <= N; i++) {
// b[i] = b[i - 1];
// b[i] += TT.query(1, 1, N, i, i).sum;
// cerr << b[i] << ' ';
// } cerr << endl;
}
// updata(1, 2, -2, 1);
for (int i = 1; i <= N; i++) {
b[i] = b[i - 1];
b[i] += TT.query(1, 1, N, i, i).sum;
// cerr << b[i] << ' ';
}
// cerr << endl;
for (int i = 1; i <= N; i++) {
int x = g[i] + 1, y = i + g[i];
if (y <= N && x <= N) {
if (x >= i) {
} else {
mp[x][a[y]] += SA.lcp(x + 1, y + 1) + 1;
}
mp[y][a[x]] += js(x, y);
// cerr << x << ' ';
// cerr << a[y] << ' ';
// cerr << y << ' ';
// cerr << a[x] << endl;
}
}
ll jg = 0;
for (int i = 1; i <= N; i++) {
ans[i] = b[i];
// int id = 0;
for (auto it : mp[i]) {
int ts = b[i] + it.second;
if (it.first == a[i]) continue;
if (ts > ans[i]) {
ans[i] = ts;
// id = it.first;
}
ans[i] = max(ans[i], ts);
// cerr << i << ' ';
// cerr << it.first << ' ';
// cerr << it.second << endl;
}
jg += (1ll * ans[i]) ^ i;
// cerr << id << ' ';
}
// cerr << endl;
// for (int i = 1; i <= N; i++) {
// cerr << b[i] << ' ';
// } cerr << endl;
// for (int i = 1; i <= N; i++) {
// cerr << ans[i] << ' ';
// }
// cerr << endl;
cout << jg << endl;
}
signed main() {
// ld be, ed;
// be = clock();
// freopen("1.in", "r", stdin);
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
srand(time(NULL));
#ifdef MULTI_CASES
int T;
cin >> T;
while (T--)
#endif
Solve();
// ed = clock();
// cerr << (ed - be) / CLOCKS_PER_SEC << endl;
}
Details
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Test #1:
score: 100
Accepted
time: 8ms
memory: 86636kb
input:
2 4 2 1 1 2 12 1 1 4 5 1 4 1 9 1 9 8 10
output:
15 217
result:
ok 2 lines
Test #2:
score: -100
Time Limit Exceeded
input:
10000 8 2 1 2 1 1 1 2 2 9 2 2 1 2 1 2 1 2 1 15 2 1 2 1 1 1 1 2 2 1 2 1 2 2 1 2 1 1 10 2 1 1 1 2 2 1 1 2 2 3 2 1 2 11 1 2 2 1 1 2 1 2 2 1 1 14 2 1 1 1 1 2 1 1 1 2 2 1 2 1 12 2 2 2 1 2 2 2 1 1 2 1 2 4 2 1 1 2 8 1 2 2 2 1 2 1 1 8 1 1 2 1 2 1 1 1 6 2 1 1 1 2 2 14 2 2 1 1 1 1 2 2 2 1 2 2 1 1 10 1 2 2 1 1...
output:
94 128 347 3 212 13 263 312 270 17 98 128 77 301 203 3 369 391 363 342 11 19 111 74 20 105 53 252 11 63 50 164 122 111 244 206 30 63 366 75 125 21 25 68 342 386 270 313 337 295 328 365 233 325 3 279 59 330 3 61 32 147 389 53 347 98 246 4 165 346 247 20 149 -4 382 400 399 73 243 348 20 54 16 327 3 9 ...