48pts数组越界求diao

#include<bits/stdc++.h>
#define ll long long
using namespace std;

const int N=1e6+5, M=N<<3;

ll n,x,x2,y,y2,cnt=0,X[N<<1];
struct Side{
	ll l,r,q,n;
}s[N<<1];
struct Node{
    ll l,r,sum,len;
}t[M];

bool cmp(Side a,Side b){
	return a.n<=b.n;
}

inline int ls(int p){return p<<1;}
inline int rs(int p){return p<<1|1;}

void push_up(int p){//Change the length of the part of tree
	if(t[p].sum)
		t[p].len=X[t[p].r+1]-X[t[p].l];
	else
		t[p].len=t[ls(p)].len+t[rs(p)].len;
}

void build(int p,int l,int r){//Easy to build a tree
    t[p]={l,r,0,0};
    if(l==r)
        return;
    int mid=(l+r)>>1;
    build(ls(p),l,mid);
    build(rs(p),mid+1,r);
}

void query(int p,int l,int r,int q){
	if(X[t[p].r+1]<=l||X[t[p].l]>=r)//The limit mods aren't in the part of tree
		return;
	if(l<=X[t[p].l]&&r>=X[t[p].r+1]){//If the limit mods are in the part of tree
		t[p].sum+=q;//Change the sum if the mods are covered
		push_up(p);//Then change the length of the part of tree
		return;//As long as get the length,stop the step
	}
	query(ls(p),l,r,q);
	query(rs(p),l,r,q);
	push_up(p);
}

int main(){
	cin>>n;
	for(int i=1;i<=n;++i){
		cin>>x>>y>>x2>>y2;
		s[++cnt].l=x;
		s[cnt].r=x2;
		s[cnt].n=y;
		s[cnt].q=1;
		X[cnt]=x;
		s[++cnt].l=x;
		s[cnt].r=x2;
		s[cnt].n=y2;
		s[cnt].q=-1;
		X[cnt]=x2;
	}
	sort(s+1,s+1+cnt,cmp);
	sort(X+1,X+1+cnt);
	unique(X+1,X+1+cnt);
	build(1,1,cnt-1);
	ll ans=0;
	for(int i=1;i<cnt;++i){
		query(1,s[i].l,s[i].r,s[i].q);
		ans+=t[1].len*(s[i+1].n-s[i].n);
	}
	cout<<ans;
	return 0;
}