LibreOJ 2018.「AHOI / HNOI2017」单旋

容易发现 Splay 最小值 / 最大值对树的影响很小且有规律可循。

然后 LCT 模拟即可(雾

亦可离散化权值后用线段树维护深度,另用一个 set 维护前驱后继。

代码:

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#include <cstdio>
#include <vector>
#include <algorithm>
#include <functional>
using namespace std;
const int N = 2e5;
const int M = 2e5;
int n,m,p1,p2;
int st[N + 5],top;
int a[N + 5],left[N + 5],right[N + 5];
vector<int> le[N + 5],ri[N + 5];
struct s_query
{
int l,r,id;
inline bool operator<(const s_query &o) const
{
return l < o.l;
}
inline bool operator>(const s_query &o) const
{
return r < o.r;
}
} qry[M + 5];
long long ans[M + 5];
struct node
{
long long sum,tag;
int ls,rs;
} seg[(N << 6) + 10];
int rt[3];
void update(int l,int r,long long k,int &p,int tl,int tr)
{
if(l > r)
return ;
static int tot = 0;
if(!p)
p = ++tot;
seg[p].sum += k * (min(r,tr) - max(l,tl) + 1);
if(l <= tl && tr <= r)
{
seg[p].tag += k;
return ;
}
int mid = tl + tr >> 1;
l <= mid && (update(l,r,k,seg[p].ls,tl,mid),1);
r > mid && (update(l,r,k,seg[p].rs,mid + 1,tr),1);
}
long long query(int l,int r,int p,int tl,int tr)
{
if(!p || (l <= tl && tr <= r))
return seg[p].sum;
int mid = tl + tr >> 1;
long long ret = seg[p].tag * (min(r,tr) - max(l,tl) + 1);
l <= mid && (ret += query(l,r,seg[p].ls,tl,mid));
r > mid && (ret += query(l,r,seg[p].rs,mid + 1,tr));
return ret;
}
int main()
{
//freopen("sf.in","r",stdin),freopen("sf.out","w",stdout);
scanf("%d%d%d%d",&n,&m,&p1,&p2);
a[0] = a[n + 1] = 0x3f3f3f3f,top = 1;
for(register int i = 1;i <= n;++i)
{
scanf("%d",a + i);
for(;top && a[st[top]] < a[i];--top);
le[left[i] = st[top]].push_back(i),st[++top] = i;
}
st[top = 1] = n + 1;
for(register int i = n;i;--i)
{
for(;top && a[st[top]] < a[i];--top);
ri[right[i] = st[top]].push_back(i),st[++top] = i;
}
for(register int i = 1;i <= m;++i)
scanf("%d%d",&qry[i].l,&qry[i].r),qry[i].id = i;
sort(qry + 1,qry + m + 1,greater<s_query>());
for(register int i = 1,j = 1;i <= m;++i)
{
for(;j <= qry[i].r;++j)
for(register int k = 0;k < ri[j].size();++k)
update(left[ri[j][k]],left[ri[j][k]],p1,rt[0],0,n + 1);
ans[qry[i].id] += query(qry[i].l,qry[i].r,rt[0],0,n + 1) + (long long)(qry[i].r - qry[i].l) * p1;
}
sort(qry + 1,qry + m + 1);
for(register int i = m,j = n;i;--i)
{
for(;j >= qry[i].l;--j)
for(register int k = 0;k < le[j].size();++k)
update(le[j][k] + 1,right[le[j][k]] - 1,p2,rt[1],0,n + 1);
ans[qry[i].id] += query(qry[i].l,qry[i].r,rt[1],0,n + 1);
}
sort(qry + 1,qry + m + 1,greater<s_query>());
for(register int i = 1,j = 1;i <= m;++i)
{
for(;j >= qry[i].l;--j)
for(register int k = 0;k < ri[j].size();++k)
update(left[ri[j][k]] + 1,ri[j][k] - 1,p2,rt[2],0,n + 1);
ans[qry[i].id] += query(qry[i].l,qry[i].r,rt[2],0,n + 1);
}
for(register int i = 1;i <= m;++i)
printf("%lld\n",ans[i]);
}