5187: [Usaco2018 Jan]Sprinklers

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
const LL mod=1e9+7;
int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||'9'<ch){if(ch=='-')f=-1;ch=getchar();}
    while('0'<=ch&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

LL a[1100000],up[1100000],lw[1100000];
LL l[1100000],suml[1100000],sumli[1100000];
int main()
{
    int n,x,y;
    n=read();
    for(int i=1;i<=n;i++)
        x=read(), y=read(), a[x]=y;
    
    //-----------------------------------------------------
    
    up[n-1]=a[n-1];for(int i=n-2;i>=0;i--)up[i]=max(up[i+1],a[i]);
    lw[0]=a[0];    for(int i=1;i<n;i++)   lw[i]=min(lw[i-1],a[i]);
    
    int now=n;
    for(int i=0;i<n;i++)
        while(now>lw[i])l[--now]=i;
    while(now>0)l[--now]=n-1;
    
    for(int i=n-1;i>=0;i--)
    {
        suml[i]=(suml[i+1]+l[i])%mod;
        sumli[i]=(sumli[i+1]+l[i]*(n-i)%mod)%mod;
    }
    
    //-----------------------------------------------------
    
    LL ans=0,d;
    for(int i=1;i<=n;i++)
    {
        d=(i*(up[i]-lw[i])*(up[i]+lw[i]+1)/2)%mod;
        ans=(ans+d)%mod;
        
        d=i*lw[i]%mod*(up[i]-lw[i])%mod;
        ans=((ans-d)%mod+mod)%mod;
        
        d=(sumli[lw[i]]-sumli[up[i]])%mod-(suml[lw[i]]-suml[up[i]])*(n-up[i])%mod;
        ans=((ans-d)%mod+mod)%mod;
    }
    printf("%lld
",ans);
    
    return 0;
}