题意:对于一段区间一共有四种操作:
题解:珂朵莉树板题
珂朵莉树,又称Old Driver Tree(ODT)。是一种基于std::set
的暴力数据结构。
关键操作:推平一段区间,使一整段区间内的东西变得一样。保证数据随机。
这道题里,这样定义珂朵莉树的节点:
struct node
{
int l,r;
mutable LL v;
node(int L, int R=-1, LL V=0):l(L), r(R), v(V) {}
bool operator<(const node& o) const
{
return l < o.l;
}
};
这样的一个节点表示[l,r]内的所有数都是v。需要注意的是mutable
,意为易变的,不定的。它对v
的修饰,使得可以在add
操作中修改v
的值。没有它的修饰会在add
函数里导致CE。
核心操作:split
#define IT set<node>::iterator
IT split(int pos)
{
IT it = s.lower_bound(node(pos));
if (it != s.end() && it->l == pos) return it;
--it;
int L = it->l, R = it->r;
LL V = it->v;
s.erase(it);
s.insert(node(L, pos-1, V));
return s.insert(node(pos, R, V)).first;
}
最后一句插入后半段,返回后半段的迭代器。这里利用了pair<iterator,bool> insert (const value_type& val)
的返回值。
推平操作:assign
void assign(int l, int r, LL val=0)
{
IT itl = split(l),itr = split(r+1);
s.erase(itl, itr);
s.insert(node(l, r, val));
}
附上代码:
#include<cstdio>
#include<set>
#include<vector>
#include<utility>
#include<algorithm>
#define IT set<node>::iterator
using std::set;
using std::vector;
using std::pair;
typedef long long LL;
const int MOD7 = 1e9 + 7;
const int MOD9 = 1e9 + 9;
const int imax_n = 1e5 + 7;
LL pow(LL a, LL b, LL mod)
{
LL res = 1;
LL ans = a % mod;
while (b)
{
if (b&1) res = res * ans % mod;
ans = ans * ans % mod;
b>>=1;
}
return res;
}
struct node
{
int l,r;
mutable LL v;
node(int L, int R=-1, LL V=0):l(L), r(R), v(V) {}
bool operator<(const node& o) const
{
return l < o.l;
}
};
set<node> s;
IT split(int pos)
{
IT it = s.lower_bound(node(pos));
if (it != s.end() && it->l == pos) return it;
--it;
int L = it->l, R = it->r;
LL V = it->v;
s.erase(it);
s.insert(node(L, pos-1, V));
return s.insert(node(pos, R, V)).first;
}
void add(int l, int r, LL val=1)
{
IT itl = split(l),itr = split(r+1);
for (; itl != itr; ++itl) itl->v += val;
}
void assign_val(int l, int r, LL val=0)
{
IT itl = split(l),itr = split(r+1);
s.erase(itl, itr);
s.insert(node(l, r, val));
}
LL rank(int l, int r, int k)
{
vector<pair<LL, int> > vp;
IT itl = split(l),itr = split(r+1);
vp.clear();
for (; itl != itr; ++itl)
vp.push_back(pair<LL,int>(itl->v, itl->r - itl->l + 1));
std::sort(vp.begin(), vp.end());
for (vector<pair<LL,int> >::iterator it=vp.begin();it!=vp.end();++it)
{
k -= it->second;
if (k <= 0) return it->first;
}
return -1LL;
}
LL sum(int l, int r, int ex, int mod)
{
IT itl = split(l),itr = split(r+1);
LL res = 0;
for (; itl != itr; ++itl)
res = (res + (LL)(itl->r - itl->l + 1) * pow(itl->v, LL(ex), LL(mod))) % mod;
return res;
}
int n, m;
LL seed, vmax;
LL rnd()
{
LL ret = seed;
seed = (seed * 7 + 13) % MOD7;
return ret;
}
LL a[imax_n];
int main()
{
scanf("%d %d %lld %lld",&n,&m,&seed,&vmax);
for (int i=1; i<=n; ++i)
{
a[i] = (rnd() % vmax) + 1;
s.insert(node(i,i,a[i]));
}
s.insert(node(n+1, n+1, 0));
int lines = 0;
for (int i =1; i <= m; ++i)
{
int op = int(rnd() % 4) + 1;
int l = int(rnd() % n) + 1;
int r = int(rnd() % n) + 1;
if (l > r)
std::swap(l,r);
int x, y;
if (op == 3)
x = int(rnd() % (r-l+1)) + 1;
else
x = int(rnd() % vmax) +1;
if (op == 4)
y = int(rnd() % vmax) + 1;
if (op == 1)
add(l, r, LL(x));
else if (op == 2)
assign_val(l, r, LL(x));
else if (op == 3)
printf("%lld\n",rank(l, r, x));
else
printf("%lld\n",sum(l, r, x, y));
}
return 0;
}