【BZOJ4241】历史研究（回滚莫队）
题目：
分析：
本校某些julao乱膜的时候发明了个“回滚邹队”，大概意思就是某个姓邹的太菜了进不了省队回滚去文化课
回滚莫队裸题qwq（话说这个名字是不是莫队本人起的啊这么萌zui
首先看到题询问区间信息+没强制在线，妥妥的莫队。然而朴素的莫队（开个桶记每种事件当前的重要度，用set或者堆之类维护一下答案）要\(O(n\sqrt n \log n)\)，直接T了……
兔崽子给我说有一种神奇的分块做法，然而我太菜了还没写，先挖个坑以后再补。
然后我去网上orz题解，看到一种叫“回滚莫队”的神奇的东西。可以发现，朴素莫队的问题在于区间变长的时候可以\(O(1)\)更新答案，但是为了维护区间变短必须要套个\(\log n\)的数据结构。所以，如果不存在区间变短的情况，就可以直接\(O(1)\)更新答案，总复杂度\(O(n\sqrt n)\)了。
考虑左端点在同一块中的所有询问，它们的右端点是单调非降的。对于左右端点在同一块中的询问，直接暴力查询即可。否则，设左端点所在块的下一块开头为\(pos\)，用一个指针\(r\)往右扫，统计\([pos,r]\)的答案。此时还有\([l,pos)\)的答案没有统计。对于每个询问，暴力\(O(\sqrt n)\)统计这部分答案，最后对两部分答案取\(max\)即可。具体参见代码。
代码：
朴素莫队（TLE）：
（一开始离散化写炸了WA了几发……菜死了）
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <queue>
#include <cmath>
#define _ 0
using namespace std;
namespace zyt
{
template<typename T>
inline bool read(T &x)
{
char c;
bool f = false;
x = 0;
do
c = getchar();
while (c != EOF && c != '-' && !isdigit(c));
if (c == EOF)
return false;
if (c == '-')
f = true, c = getchar();
do
x = x * 10 + c - '0', c = getchar();
while (isdigit(c));
if (f)
x = -x;
return true;
}
template<typename T>
inline void write(T x)
{
char buf[20];
char *pos = buf;
if (x < 0)
putchar('-'), x = -x;
do
*pos++ = x % 10 + '0';
while (x /= 10);
while (pos > buf)
putchar(*--pos);
}
typedef long long ll;
const int N = 1e5 + 10, Q = 1e5 + 10;
int n, q, block, belong[N], arr[N], tmp[N];
ll ans[N];
struct _ask
{
int l, r, id;
bool operator < (const _ask &b) const
{
return belong[l] == belong[b.l] ? r < b.r : belong[l] < belong[b.l];
}
}ask[Q];
namespace Mo_Algorithm
{
priority_queue<ll> pq, del;
ll num[N];
void update(const int pos, const int x)
{
del.push(num[pos]);
num[pos] += (ll)tmp[pos] * x;
pq.push(num[pos]);
while (!del.empty() && pq.top() == del.top())
pq.pop(), del.pop();
}
void solve(const int maxx)
{
int l = 1, r = 1;
for (int i = 1; i <= maxx; i++)
pq.push(0);
update(arr[1], 1);
for (int i = 1; i <= q; i++)
{
while (l < ask[i].l)
update(arr[l++], -1);
while (l > ask[i].l)
update(arr[--l], 1);
while (r < ask[i].r)
update(arr[++r], 1);
while (r > ask[i].r)
update(arr[r--], -1);
ans[ask[i].id] = pq.top();
}
}
}
int work()
{
read(n), read(q);
block = pow(n, 0.5);
for (int i = 1; i <= n; i++)
{
read(arr[i]), tmp[i] = arr[i];
belong[i] = (i - 1) / block + 1;
}
sort(tmp + 1, tmp + n + 1);
int cnt = unique(tmp + 1, tmp + n + 1) - tmp;
for (int i = 1; i <= n; i++)
arr[i] = lower_bound(tmp + 1, tmp + cnt, arr[i]) - tmp;
for (int i = 1; i <= q; i++)
{
read(ask[i].l), read(ask[i].r);
ask[i].id = i;
}
sort(ask + 1, ask + q + 1);
Mo_Algorithm::solve(cnt);
for (int i = 1; i <= q; i++)
write(ans[i]), putchar('\n');
return (0^_^0);
}
}
int main()
{
return zyt::work();
}
回滚莫队（AC）：
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <queue>
#include <cmath>
#define _ 0
using namespace std;
namespace zyt
{
template<typename T>
inline bool read(T &x)
{
char c;
bool f = false;
x = 0;
do
c = getchar();
while (c != EOF && c != '-' && !isdigit(c));
if (c == EOF)
return false;
if (c == '-')
f = true, c = getchar();
do
x = x * 10 + c - '0', c = getchar();
while (isdigit(c));
if (f)
x = -x;
return true;
}
template<typename T>
inline void write(T x)
{
char buf[20];
char *pos = buf;
if (x < 0)
putchar('-'), x = -x;
do
*pos++ = x % 10 + '0';
while (x /= 10);
while (pos > buf)
putchar(*--pos);
}
typedef long long ll;
const int N = 1e5 + 10, Q = 1e5 + 10;
int n, q, block, belong[N], arr[N], tmp[N];
ll ans[N];
struct _ask
{
int l, r, id;
bool operator < (const _ask &b) const
{
return belong[l] == belong[b.l] ? r < b.r : belong[l] < belong[b.l];
}
}ask[Q];
inline int begin(const int a)
{
return (a - 1) * block + 1;
}
namespace Mo_Algorithm
{
ll solve_small(const int l, const int r)
{
static ll num[N];
ll ans = 0;
for (int i = l; i <= r; i++)
ans = max(ans, num[arr[i]] += tmp[arr[i]]);
for (int i = l; i <= r; i++)
num[arr[i]] = 0;
return ans;
}
void solve(const int maxx)
{
static ll num[N];
ll now = 0;
int l, r, lbegin;
for (int i = 1; i <= q; i++)
{
if (belong[ask[i].l] != belong[ask[i - 1].l])
{
memset(num, 0, sizeof(ll[maxx + 1]));
lbegin = begin(belong[ask[i].l] + 1), l = lbegin, r = lbegin - 1;
now = 0;
}
if (belong[ask[i].l] == belong[ask[i].r])
ans[ask[i].id] = solve_small(ask[i].l, ask[i].r);
else
{
while (r < ask[i].r)
{
++r;
now = max(now, num[arr[r]] += tmp[arr[r]]);
}
ll bck = now;
while (l > ask[i].l)
{
--l;
now = max(now, num[arr[l]] += tmp[arr[l]]);
}
ans[ask[i].id] = now;
while (l < lbegin)
{
num[arr[l]] -= tmp[arr[l]];
++l;
}
now = bck;
}
}
}
}
int work()
{
read(n), read(q);
block = pow(n, 0.5);
for (int i = 1; i <= n; i++)
{
read(arr[i]), tmp[i] = arr[i];
belong[i] = (i - 1) / block + 1;
}
sort(tmp + 1, tmp + n + 1);
int cnt = unique(tmp + 1, tmp + n + 1) - tmp;
for (int i = 1; i <= n; i++)
arr[i] = lower_bound(tmp + 1, tmp + cnt, arr[i]) - tmp;
for (int i = 1; i <= q; i++)
{
read(ask[i].l), read(ask[i].r);
ask[i].id = i;
}
sort(ask + 1, ask + q + 1);
Mo_Algorithm::solve(cnt);
for (int i = 1; i <= q; i++)
write(ans[i]), putchar('\n');
return (0^_^0);
}
}
int main()
{
return zyt::work();
}
