【CF526F】Pudding Monsters cdq分治

【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。

$n\le 3\times 10^5$

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