【CF526F】Pudding Monsters
题意：给你一个排列$p_i$，问你有对少个区间的值域段是连续的。
$nle 3 imes 10^5$
题解：bzoj3745 Norma 的弱化版。直接cdq分治，考虑最大值和最小值分别在左右两边的情况。这里就当练练手了。
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=300010;
typedef long long ll;
const int inf=1<<30;
ll ans;
int n;
int v[maxn],rm[maxn],rn[maxn],lm[maxn],ln[maxn],s1[maxn<<1],s2[maxn<<1];
inline int rd()
{
int ret=0,f=1; char gc=getchar();
while(gc<'0'||gc>'9') {if(gc=='-') f=-f; gc=getchar();}
while(gc>='0'&&gc<='9') ret=ret*10+(gc^'0'),gc=getchar();
return ret*f;
}
void solve(int l,int r)
{
if(l==r) return ;
int i,j1,j2,mid=(l+r)>>1;
solve(l,mid),solve(mid+1,r);
for(lm[mid+1]=0,ln[mid+1]=inf,i=mid;i>=l;i--) lm[i]=max(lm[i+1],v[i]),ln[i]=min(ln[i+1],v[i]);
for(rm[mid]=0,rn[mid]=inf,i=mid+1;i<=r;i++) rm[i]=max(rm[i-1],v[i]),rn[i]=min(rn[i-1],v[i]);
for(i=mid,j1=j2=mid+1;i>=l;i--)
{
while(j1<=r&&rn[j1]>ln[i]&&rm[j1]<lm[i]) s1[rn[j1]+j1]++,j1++;
while(j2<=r&&rm[j2]<lm[i]) s2[rn[j2]+j2]++,j2++;
if(j1!=mid+1&&j1-1>=i+lm[i]-ln[i]) ans++;
ans+=s2[i+lm[i]]-s1[i+lm[i]];
}
for(i=mid+1;i<=r;i++) s1[rn[i]+i]=s2[rn[i]+i]=0;
for(i=mid+1,j1=j2=mid;i<=r;i++)
{
while(j1>=l&&ln[j1]>rn[i]&&lm[j1]<rm[i]) s1[ln[j1]-j1+n]++,j1--;
while(j2>=l&&lm[j2]<rm[i]) s2[ln[j2]-j2+n]++,j2--;
if(j1!=mid&&j1+1<=i-(rm[i]-rn[i])) ans++;
ans+=s2[rm[i]-i+n]-s1[rm[i]-i+n];
}
for(i=mid;i>=l;i--) s1[ln[i]-i+n]=s2[ln[i]-i+n]=0;
}
int main()
{
n=rd();
int i,a;
for(i=1;i<=n;i++) a=rd(),v[a]=rd();
solve(1,n);
printf("%lld",ans+n);
return 0;
}//3 1 1 2 2 3 3