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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#604082 | #7685. Barkley II | longyin | TL | 4ms | 19536kb | C++17 | 5.3kb | 2024-10-01 22:57:49 | 2024-10-01 22:57:50 |
Judging History
answer
#include <bits/stdc++.h>
#define maxn 1000007
#define int long long
#define dl long double
#define mod 1000000007
#define inf 0x3f3f3f3f
#pragma GCC optimize(2)
using namespace std;
int n, m;
inline int pls(int a, int b) {int m = a + b; return m < mod ? m : m - mod;}
inline int dec(int a, int b) {int m = a - b; return m < 0 ? m + mod : m;}
inline int mul(int a, int b) {return 1ll * a * b % mod;}
inline int fpow(int a, int b) {
int ans = 1;
for(; b; b >>= 1,a = mul(a, a)) if(b & 1) ans = mul(ans, a);
return ans;
}
inline int inv(int a) {return fpow(a, mod - 2);}
inline int dvi(int a, int b) {return mul(a, inv(b));};
inline int qread() {
char c = getchar(); int num = 0, f = 1;
for(; !isdigit(c); c=getchar()) if(c == '-') f = -1;
for(; isdigit(c); c=getchar()) num = num * 10 + c - '0';
return num * f;
}
#define ls(u) tr[u].s[0]
#define rs(u) tr[u].s[1]
int read() {
int res = 0, flg = 1; char c = getchar();
while(c != '-' && !isdigit(c)) c = getchar();
if(c == '-') flg = 0, c = getchar();
while(isdigit(c)) {
res = (res << 3) + (res << 1) + (c ^ 48);
c = getchar();
}
return flg ? res : -res;
}
const int N = 2e5 + 10;
struct Node{
int s[2];
int sum;
}tr[N * ((int)log2(N) + 1) * 4];
int tot, root[N];
void update(int u) {
tr[u].sum = tr[ls(u)].sum + tr[rs(u)].sum;
}
void build(int& u, int l, int r) {
u = ++tot;
if(l == r) return;
int mid = (l + r) >> 1;
build(ls(u), l, mid);
build(rs(u), mid + 1, r);
}
void add(int u, int& v, int l, int r, int val) {
int mid = (l + r) >> 1;
v = ++tot;
tr[v] = {0};
if(l == r) {
tr[v].sum = 1;
return;
}
if(val <= mid) {
rs(v) = rs(u);
add(ls(u), ls(v), l, mid, val);
} else {
ls(v) = ls(u);
add(rs(u), rs(v), mid + 1, r, val);
}
update(v);
}
int check_sum(int u, int v, int l, int r, int ll, int rr) {
if(rr < ll) return 0;
if(ll <= l && rr >= r) return tr[v].sum - tr[u].sum;
int mid = (l + r) >> 1;
int res = 0;
if(ll <= mid) res += check_sum(ls(u), ls(v), l, mid, ll, rr);
if(rr > mid) res += check_sum(rs(u), rs(v), mid + 1, r, ll, rr);
return res;
}
int check_w(int u, int v, int l, int r) {
if(tr[v].sum - tr[u].sum == r - l + 1) return 0;
if(l == r) return l;
int res = 0;
int mid = (l + r) >> 1;
res = check_w(ls(u), ls(v), l, mid);
if(res) return res;
return check_w(rs(u), rs(v), mid + 1, r);
}
int check(int l, int r) {
//cout << l << ' ' << r << '\n';
if(l > r) return 0;
return tr[root[r]].sum - tr[root[l - 1]].sum - check_w(root[l - 1], root[r], 1, 2 * n);
}
int val[N];
vector<int> f, bd[N];
tuple<int, int, int> query[N];
int qcnt;
int ans[N];
int vis[N];
struct BIT {
int tr[N];
int n;
BIT(int n) : n(n) {
for(int i = 1; i <= n; i++) {
tr[i] = 0;
}
}
int lowbit(int x) {
return x & -x;
}
void add(int x, int k) {
for (int i = x; i <= n; i += lowbit(i)) {
tr[i] += k;
}
}
long long sum(int x) {
long long res = 0;
for (int i = x; i >= 1; i -= lowbit(i)) {
res += tr[i];
}
return res;
}
};
void cal() {
sort(query + 1, query + qcnt + 1);
BIT bit(n);
for (int i = 1; i <= n; i++) {
vis[i] = 0;
ans[i] = 0;
}
for (int i = 1, start = 1; i <= qcnt; i++) {
int l = get<0>(query[i]);
int r = get<1>(query[i]);
int id = get<2>(query[i]);
for (int j = start; j <= r; j++) {
if (vis[val[j]])
bit.add(vis[val[j]], -1);
bit.add(j, 1);
vis[val[j]] = j;
}
start = r + 1;
ans[id] = bit.sum(r) - bit.sum(l - 1);
}
}
void solve() {
n = read(), m = read();
qcnt = tot = 0;
build(root[0], 1, 2 * n);
for(int i = 1; i <= n; i++) {
val[i] = read();
f.emplace_back(val[i]);
f.emplace_back(i);
}
sort(f.begin(), f.end());
f.erase(unique(f.begin(), f.end()), f.end());
for(int i = 1; i <= n; i++) {
add(root[i - 1], root[i], 1, 2 * n, upper_bound(f.begin(), f.end(), val[i]) - f.begin());
bd[upper_bound(f.begin(), f.end(), val[i]) - f.begin()].push_back(i);
}
for(int i = 1; i <= 2 * n; i++) {
if(!bd[i].size()) continue;
query[++qcnt] = tuple<int, int, int>(1, bd[i][0] - 1, qcnt);
for(int j = 1; j < bd[i].size(); j++) {
query[++qcnt] = tuple<int, int, int>(bd[i][j - 1] + 1, bd[i][j] - 1, qcnt);
}
query[++qcnt] = tuple<int, int, int>(bd[i][bd[i].size() - 1] + 1, n, qcnt);
}
cal();
int res = check(1, n);
for (int i = 1; i <= qcnt; i++) {
int l = get<0>(query[i]);
int r = get<1>(query[i]);
int id = get<2>(query[i]);
if(l > r) continue;
int num = ans[id];
//cout << num << ' ' << l << ' ' << r << "*\n";
res = max(res, num - check_w(root[l - 1], root[r], 1, 2 * n));
}
cout << res << '\n';
/*int res = check(1, n);
for(int i = 1; i <= 2 * n; i++) {
if(!bd[i].size()) continue;
res = max(res, check(1, bd[i][0] - 1));
for(int j = 1; j < bd[i].size(); j++) {
res = max(res, check(bd[i][j - 1] + 1, bd[i][j] - 1));
}
res = max(res, check(bd[i][bd[i].size() - 1] + 1, n));
}
cout << res << '\n'; */
}
signed main() {
int t = qread();
while(t--) {
solve();
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 4ms
memory: 19536kb
input:
2 5 4 1 2 2 3 4 5 10000 5 2 3 4 1
output:
2 3
result:
ok 2 number(s): "2 3"
Test #2:
score: -100
Time Limit Exceeded
input:
50000 10 19 12 6 1 12 11 15 4 1 13 18 10 8 8 7 6 7 6 2 2 3 4 8 10 6 3 2 6 6 5 2 3 4 5 6 10 11 6 3 7 9 2 1 2 10 10 4 10 6 6 1 2 6 1 1 3 4 2 1 10 9 8 5 3 9 1 7 5 5 1 1 10 5 1 4 3 2 5 4 5 3 5 2 10 14 3 8 12 10 4 2 3 13 7 3 10 14 5 5 12 2 8 1 13 9 8 5 10 7 5 5 6 6 1 5 3 7 3 4 10 7 5 1 4 6 1 6 4 3 7 5 10...
output:
6 5 4 4 4 4 3 7 4 4 5 5 2 3 6 6 7 5 7 5 5 5 6 3 6 8 7 2 5 5 5 2 4 4 4 5 5 3 7 3 3 4 6 4 6 5 4 4 3 8 6 6 5 7 5 4 5 4 6 6 6 3 7 3 6 4 3 7 6 6 6 7 4 3 6 4 4 5 4 5 6 6 4 5 5 9 4 5 7 3 6 5 2 2 5 6 6 8 4 3 4 5 5 5 7 7 3 2 6 5 3 4 4 4 5 6 5 5 6 7 7 5 5 7 4 7 3 7 6 6 6 5 4 4 5 4 3 3 6 5 4 6 5 5 5 3 5 6 5 6 ...