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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 |
---|---|---|---|---|---|---|---|---|---|
#383285 | #5069. Vacation | hhoppitree | WA | 3ms | 20056kb | C++14 | 9.9kb | 2024-04-09 09:23:44 | 2024-04-09 09:23:45 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
long long States[N * 20], *nowState = States;
inline long long* myMalloc(int sz, int flg = 0)
{
long long *sta = nowState;
nowState += sz;
if (flg) {
fill(sta, nowState, (long long)-1e18);
}
return sta;
}
char I[40000050], *J = I, O[8000050], *o = O;
inline int read()
{
unsigned int x = 0;
bool zf = 0;
while ((*J < 48 || 57 < *J) && (*J) != '-') ++J;
((*J++ == '-') ? (zf = 1) : x = *(J - 1) ^ 48);
while (47 < *J && *J < 58) x = (x << 1) + (x << 3) + (*J++ ^ 48);
return (zf ? -(int)x : x);
}
inline void print(unsigned long long x)
{
static unsigned long long S[16], T = 0, y;
do y = x / 10, S[T++] = x - y * 10; while(x = y);
while (T) *o++ = S[--T] ^ 48;
}
int n, m, C;
long long a[N];
namespace SEG1
{
typedef long long LL;
typedef tuple<LL, LL, LL, LL> dt;
int sz;
dt z[1 << 22];
inline dt operator + (dt x, dt y)
{
auto [a, b, c, d] = x;
auto [e, f, g, h] = y;
return {a + e, max({b, f, d + g}), max(c, a + g), max(h, e + d)};
}
inline void build()
{
sz = 1;
while (sz <= n + 1) {
sz <<= 1;
}
for (int i = 1; i <= n; ++i) {
z[i + sz] = {a[i], max(a[i], 0ll), max(a[i], 0ll), max(a[i], 0ll)};
}
for (int i = (n + sz) >> 1; i; --i) {
z[i] = z[i << 1] + z[i << 1 | 1];
}
return;
}
inline void modify(int x)
{
z[x + sz] = {a[x], max(a[x], 0ll), max(a[x], 0ll), max(a[x], 0ll)};
x += sz;
while (x >>= 1) {
z[x] = z[x << 1] + z[x << 1 | 1];
}
return;
}
inline long long query(int L, int R)
{
dt rL = {0, 0, 0, 0}, rR = {0, 0, 0, 0};
for (L += sz - 1, R += sz + 1; L ^ R ^ 1; L >>= 1, R >>= 1) {
(!(L & 1)) && (rL = rL + z[L ^ 1], 0);
(R & 1) && (rR = z[R ^ 1] + rR, 0);
}
auto [A, B, C, D] = rL + rR;
return B;
}
inline long long cb(dt z)
{
auto &[A, B, C, D] = z;
return A;
}
inline long long querySum(int L, int R)
{
if (L > R) {
return 0ll;
}
long long S = 0;
for (L += sz - 1, R += sz + 1; L ^ R ^ 1; L >>= 1, R >>= 1) {
(!(L & 1)) && (S += cb(z[L ^ 1]), 0);
(R & 1) && (S += cb(z[R ^ 1]), 0);
}
return S;
}
}
int bl;
long long glo1[N], glo2[N];
namespace SEG2
{
int n, sz;
long long mx[1 << 22];
inline void build()
{
n = bl - 2, sz = 1;
while (sz <= n + 1) {
sz <<= 1;
}
for (int i = 1; i <= n; ++i) {
mx[i + sz] = max(glo1[i], glo2[i]);
}
for (int i = (n + sz) >> 1; i; --i) {
mx[i] = max(mx[i << 1], mx[i << 1 | 1]);
}
return;
}
inline void modify(int x)
{
mx[x + sz] = max(glo1[x], glo2[x]);
x += sz;
while (x >>= 1) {
mx[x] = max(mx[x << 1], mx[x << 1 | 1]);
}
return;
}
inline long long query(int L, int R)
{
long long res = 0;
for (L += sz - 1, R += sz + 1; L ^ R ^ 1; L >>= 1, R >>= 1) {
(!(L & 1)) && (res = max(res, mx[L ^ 1]));
((R & 1)) && (res = max(res, mx[R ^ 1]));
}
return res;
}
}
long long Sa[N], Sb[N];
inline tuple<long long, long long, long long, long long, long long> operator + (tuple<long long, long long, long long, long long, long long> x, tuple<long long, long long, long long, long long, long long> y);
struct DS
{
int len, sz;
typedef long long LL;
LL *SuA, *SuB, *MxA, *MxB, *S;
friend inline tuple<LL, LL, LL, LL, LL> operator + (tuple<LL, LL, LL, LL, LL> x, tuple<LL, LL, LL, LL, LL> y)
{
auto &[A, B, C, D, E] = x;
auto &[F, G, H, I, J] = y;
return {A + F, B + G, max(H, C + F), max(D, I + B), max({E + F, J + B, H + D})};
}
inline void Build()
{
int t = 2 * len, z = 1;
while (z <= t + 1) {
z <<= 1;
}
sz = z;
SuA = myMalloc(len + sz + 1);
SuB = myMalloc(len + sz + 1);
MxA = myMalloc(len + sz + 1, 1);
MxB = myMalloc(len + sz + 1, 1);
S = myMalloc(len + z + 1, 1);
for (int i = 1; i <= len; ++i) {
SuA[i + sz] = MxA[i + sz] = Sa[i];
SuB[i + sz] = MxB[i + sz] = Sb[i];
S[i + sz] = -1e18;
}
for (int k = (len + z) >> 1; k; --k) {
SuA[k] = SuA[k << 1] + SuA[k << 1 | 1];
SuB[k] = SuB[k << 1] + SuB[k << 1 | 1];
MxA[k] = max(MxA[k << 1 | 1], SuA[k << 1 | 1] + MxA[k << 1]);
MxB[k] = max(MxB[k << 1], SuB[k << 1] + MxB[k << 1 | 1]);
S[k] = max({S[k << 1] + SuA[k << 1 | 1], S[k << 1 | 1] + SuB[k << 1], MxA[k << 1 | 1] + MxB[k << 1]});
}
return;
}
inline long long query(int L, int R)
{
tuple<LL, LL, LL, LL, LL> tA, tB;
for (L += sz - 1, R += sz + 1; L ^ R ^ 1; L >>= 1, R >>= 1) {
(!(L & 1)) && (tA = tA + tuple<LL, LL, LL, LL, LL>{SuA[L ^ 1], SuB[L ^ 1], MxA[L ^ 1], MxB[L ^ 1], S[L ^ 1]}, 0);
((R & 1)) && (tB = tuple<LL, LL, LL, LL, LL>{SuA[R ^ 1], SuB[R ^ 1], MxA[R ^ 1], MxB[R ^ 1], S[R ^ 1]} + tB, 0);
}
auto [A, B, C, D, E] = tA + tB;
return E;
}
inline void modifyA(int k, int y)
{
k += sz;
SuA[k] += y, MxA[k] += y;
while (k >>= 1) {
SuA[k] = SuA[k << 1] + SuA[k << 1 | 1];
MxA[k] = max(MxA[k << 1 | 1], SuA[k << 1 | 1] + MxA[k << 1]);
S[k] = max({S[k << 1] + SuA[k << 1 | 1], S[k << 1 | 1] + SuB[k << 1], MxA[k << 1 | 1] + MxB[k << 1]});
}
return;
}
inline void modifyB(int k, int y)
{
k += sz;
SuB[k] += y, MxB[k] += y;
while (k >>= 1) {
SuB[k] = SuB[k << 1] + SuB[k << 1 | 1];
MxB[k] = max(MxB[k << 1], SuB[k << 1] + MxB[k << 1 | 1]);
S[k] = max({S[k << 1] + SuA[k << 1 | 1], S[k << 1 | 1] + SuB[k << 1], MxA[k << 1 | 1] + MxB[k << 1]});
}
return;
}
inline void modify(int type, int x, int y)
{
if (!type) {
modifyA(x, y);
} else {
modifyB(x, y);
}
return;
}
} SEG3[N];
inline long long calc(int wh, int L = 0, int R = 0, long long V = 0)
{
if (!L) {
return SEG3[wh].S[1];
}
if (SEG1::query(L, R + C) <= V) {
return V;
}
L -= (wh - 1) * C, R -= (wh - 1) * C;
return SEG3[wh].query(L, R) + SEG1::querySum(wh * C + 1, wh * C + L - 1) + SEG1::querySum((wh - 1) * C + R + 1, wh * C);
}
signed main()
{
fread(I, 1, 40000038, stdin);
n = read(), m = read(), C = read();
bl = (n - 1) / C + 1;
for (int i = 1; i <= n; ++i) {
a[i] = read();
}
SEG1::build();
for (int i = 2; i <= bl - 1; ++i) {
long long s = 0, mx = 0;
for (int j = (i - 1) * C + 1; j <= i * C && j <= n; ++j) {
mx = max(mx, s = max(s, 0ll) + a[j]);
}
glo1[i] = mx;
}
for (int i = 1; i <= bl - 1; ++i) {
SEG3[i].len = min((i + 1) * C, n) - i * C;
for (int j = 1; j <= SEG3[i].len; ++j) {
Sa[j] = 0;
}
for (int j = 1; j <= C && j + (i - 1) * C <= n; ++j) {
Sa[min(j, SEG3[i].len)] += a[j + (i - 1) * C];
}
for (int j = 1; j <= SEG3[i].len; ++j) {
Sb[j] = a[j + i * C];
}
SEG3[i].Build();
}
for (int i = 2; i <= bl - 2; ++i) {
glo2[i] = calc(i);
}
SEG2::build();
while (m--) {
int opt = read(), L = read(), R = read();
if (opt == 1) {
if (a[L] == R) {
continue;
}
int D = R - a[L];
a[L] = R;
SEG1::modify(L);
int bel = (L - 1) / C + 1;
long long flg = max(glo1[bel], glo2[bel]);
if (bel >= 2 && bel <= bl - 1) {
glo1[bel] = SEG1::query((bel - 1) * C + 1, bel * C);
}
if (bel >= 2) {
SEG3[bel - 1].modify(1, L - (bel - 1) * C, D);
}
if (bel <= bl - 1) {
SEG3[bel].modify(0, L - (bel - 1) * C, D);
}
if (bel >= 3) {
long long tv = max(glo1[bel - 1], glo2[bel - 1]);
glo2[bel - 1] = calc(bel - 1);
if (max(glo1[bel - 1], glo2[bel - 1]) != tv && bel <= bl - 1) {
SEG2::modify(bel - 1);
}
}
if (bel >= 2 && bel <= bl - 2) {
glo2[bel] = calc(bel);
}
if (bel >= 2 && bel <= bl - 2 && flg != max(glo1[bel], glo2[bel])) {
SEG2::modify(bel);
}
} else {
if (R - L + 1 <= C) {
print(SEG1::query(L, R));
*o++ = '\n';
continue;
}
int bel = (L - 1) / C + 1, ber = (R - 1) / C + 1;
long long res = (bel + 2 >= ber ? 0ll : SEG2::query(bel + 1, ber - 2));
if (bel + 1 != ber) {
res = max(res, glo1[ber - 1]);
}
res = max({res, SEG1::query(L, L + C - 1), SEG1::query(R - C + 1, R)});
if (ber == bel + 1) {
res = max(res, calc(bel, L, R - C, res));
} else {
res = max({res, calc(bel, L, bel * C, res), calc(ber - 1, (ber - 2) * C + 1, R - C, res)});
}
print(res);
*o++ = '\n';
}
}
fwrite(O, 1, o - O, stdout);
return 0;
}
詳細信息
Test #1:
score: 0
Wrong Answer
time: 3ms
memory: 20056kb
input:
5 6 3 0 -5 -3 8 -3 2 3 5 1 2 5 2 1 5 1 4 -3 2 3 5 2 1 5
output:
8 8 0 5
result:
wrong answer 2nd numbers differ - expected: '10', found: '8'