P4690 [Ynoi2016] 镜中的昆虫
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P4690 [Ynoi2016] 镜中的昆虫
区间赋值区间数颜色，\(n \leq 10^5\)，值域 \([1,10^9]\)，要求线性空间。
sol
首先考虑经典数颜色套路，设 \(pre_i\) 表示上一个与 \(a_i\) 相同的数的位置。
对于区间赋值操作，我们发现性质：\(\forall i\in(l,r],pre_i ←i-1\)，对于 \(i=l\) 或区间外的情况特判即可。
考虑使用 ODT 维护 \(pre\) 的修改，一个 ODT 维护序列，一个 ODT 数组维护颜色位置，均摊修改次数为 \(\mathcal O(n)\)（增加节点总数 \(\mathcal O(n)\)，删除节点总数 \(\mathcal O(n)\)），时间复杂度 \(\mathcal O(n\log n)\)。
那么问题转化为带修求 \(\sum\limits_{i=l}^{r}[pre_i<l]\)，考虑到要求线性空间，故使用三维 cdq，时间复杂度 \(\mathcal O(n \log^2 n)\)。
时间复杂度 \(\mathcal O(n \log ^2 n)\)，空间复杂度 \(\mathcal O(n)\)。
详见代码。
#include <bits/stdc++.h>
using namespace std;
namespace Fread
{
const int SIZE = 1 << 21;
char buf[SIZE], *S, *T;
inline char getchar()
{
if (S == T)
{
T = (S = buf) + fread(buf, 1, SIZE, stdin);
if (S == T)
return '\n';
}
return *S++;
}
}
namespace Fwrite
{
const int SIZE = 1 << 21;
char buf[SIZE], *S = buf, *T = buf + SIZE;
inline void flush()
{
fwrite(buf, 1, S - buf, stdout);
S = buf;
}
inline void putchar(char c)
{
*S++ = c;
if (S == T)
flush();
}
struct NTR
{
~NTR()
{
flush();
}
} ztr;
}
#ifdef ONLINE_JUDGE
#define getchar Fread::getchar
#define putchar Fwrite::putchar
#endif
inline int read()
{
int x = 0, f = 1;
char c = getchar();
while(c < '0' || c > '9')
{
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
{
x = (x << 3) + (x << 1) + c - 48;
c = getchar();
}
return x * f;
}
inline void write(int x)
{
if(x < 0) putchar('-'), x = -x;
if(x > 9) write(x / 10);
putchar(x % 10 + 48);
}
const int _ = 2e5 + 10, M = 1e6 + 10;
int n, m, tot, a[_], cnt, b[_], ans[_], d[_], pre[_];
struct Query
{
int opt, l, r, x;
} q[_];
struct Node
{
int x, y, z, k, d, ans;
inline bool operator < (const Node &t) const
{
return x != t.x ? x < t.x : (y != t.y ? y < t.y : z < t.z);
}
} p[M], c[M >> 1];
#define pii pair<int, int>
#define pa set<pii>::iterator
struct Col
{
set<pii> cl, v[_];
inline void split(int k)
{
pa it = cl.lower_bound({k + 1, 0});
int r = it->first - 1;
--it;
int l = it->first;
if(k == r) return;
int c = it->second;
cl.erase(it);
cl.insert({l, c});
cl.insert({k + 1, c});
v[c].erase({l, r});
v[c].insert({l, k});
v[c].insert({k + 1, r});
}
inline void Insert(int l, int r, int c, int t)
{
split(l - 1), split(r);
for(pa it = cl.lower_bound({l, 0}); it-> first <= r;)
{
int x = it->first, y = it->second;
++it;
int z = it->first - 1;
cl.erase({x, y});
v[y].erase({x, z});
if(x == l)
{
int e = (--(v[c].upper_bound({x, 0})))->second;
if(pre[x] != e)
{
p[++tot] = {t, x, pre[x], 0, -1, 0};
pre[x] = e;
p[++tot] = {t, x, pre[x], 0, 1, 0};
}
}
else
{
if(pre[x] != x - 1)
{
p[++tot] = {t, x, pre[x], 0, -1, 0};
pre[x] = x - 1;
p[++tot] = {t, x, pre[x], 0, 1, 0};
}
}
x = z + 1;
z = (v[y].lower_bound({x, 0}))->first;
int e = (--v[y].lower_bound({l, 0}))->second;
if(z > r && z <= n && pre[z] != e)
{
p[++tot] = {t, z, pre[z], 0, -1, 0};
pre[z] = e;
p[++tot] = {t, z, pre[z], 0, 1, 0};
}
}
pa it = v[c].lower_bound({r + 1, 0});
int x = it->first;
if(x <= n && pre[x] != r)
{
p[++tot] = {t, x, pre[x], 0, -1, 0};
pre[x] = r;
p[++tot] = {t, x, pre[x], 0, 1, 0};
}
v[c].insert({l, r});
cl.insert({l, c});
x = r + 1;
if(x <= n)
{
it = cl.lower_bound({x, 0});
int y = it->second;
int e = (--v[y].lower_bound({x, 0}))->second;
if(pre[x] != e)
{
p[++tot] = {t, x, pre[x], 0, -1, 0};
pre[x] = e;
p[++tot] = {t, x, pre[x], 0, 1, 0};
}
}
}
} col;
inline void update(int x, int v)
{
++x;
while(x < _) d[x] += v, x += x & -x;
}
inline int query(int x)
{
++x;
int res = 0;
while(x) res += d[x], x -= x & -x;
return res;
}
void solve(int l, int r)
{
if(l >= r) return;
int mid = (l + r) >> 1;
solve(l, mid), solve(mid + 1, r);
int i = l, j = mid + 1, k = 0;
bool flg = (l != 1 || r != tot);
while(i <= mid && j <= r)
if(p[i].y <= p[j].y)
{
if(p[i].k == 0) update(p[i].z, p[i].d);
if(flg) c[k++] = p[i];
++i;
}
else
{
if(p[j].k == 1) p[j].ans += query(p[j].z);
if(flg) c[k++] = p[j];
++j;
}
while(i <= mid)
{
if(p[i].k == 0) update(p[i].z, p[i].d);
if(flg) c[k++] = p[i];
++i;
}
while(j <= r)
{
if(p[j].k == 1) p[j].ans += query(p[j].z);
if(flg) c[k++] = p[j];
++j;
}
for(int i = l; i <= mid; ++i) if(p[i].k == 0) update(p[i].z, -p[i].d);
if(flg) for(int i = 0; i < k; ++i) p[i + l] = c[i];
}
signed main()
{
n = cnt = read(), m = read();
for(int i = 1; i <= n; ++i) a[i] = b[i] = read();
for(int i = 1; i <= m; ++i)
{
q[i].opt = read(), q[i].l = read(), q[i].r = read();
if(q[i].opt == 1) b[++cnt] = q[i].x = read();
}
sort(b + 1, b + cnt + 1);
cnt = unique(b + 1, b + cnt + 1) - b - 1;
for(int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + cnt + 1, a[i]) - b;
for(int i = 1; i <= m; ++i) if(q[i].opt == 1) q[i].x = lower_bound(b + 1, b + cnt + 1, q[i].x) - b;
for(int i = 1; i <= cnt; ++i) col.v[i].insert({0, 0});
for(int i = 1; i <= n; ++i)
{
pre[i] = (--col.v[a[i]].end())->second;
p[++tot] = {0, i, pre[i], 0, 1, 0};
col.v[a[i]].insert({i, i});
col.cl.insert({i, a[i]});
}
for(int i = 1; i <= cnt; ++i) col.v[i].insert({n + 1, 0});
col.cl.insert({0, 0}), col.cl.insert({n + 1, n + 1});
for(int i = 1; i <= m; ++i)
if(q[i].opt == 1) col.Insert(q[i].l, q[i].r, q[i].x, i);
else p[++tot] = {i, q[i].r, q[i].l - 1, 1, 1, 0}, p[++tot] = {i, q[i].l - 1, q[i].l - 1, 1, -1, 0};
sort(p + 1, p + tot + 1);
solve(1, tot);
for(int i = 1; i <= tot; ++i) if(p[i].k) ans[p[i].x] += p[i].ans * p[i].d;
for(int i = 1; i <= m; ++i) if(q[i].opt == 2) write(ans[i]), putchar('\n');
return 0;
}
标签：pre,cl,int,Ynoi2016,P4690,++,second,tot,镜中 来源： https://www.cnblogs.com/orzz/p/16338943.html