P5111 zhtobu3232的线段树

传送门

用一个扫描线的方法刚了俩小时最后发现空间开不下……
顺便吐槽一句这个线段树的写法一点都不优美……
题解

//minamoto
#include<bits/stdc++.h>
#define R register
#define ll long long
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
ll read(){
    R ll res,f=1;R char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
    return res*f;
}
const int N=2e7+5,P=998244353;
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
int ls[N],rs[N],Pre[N],suf[N];bool vis[N];
ll n,l,r;int m,ans,cnt,rt;
inline int calc(R int x){return 1ll*x*(x+1)/2%P;}
void build(int &p,ll l,ll r,ll ql,ll qr){
    if(!p)p=++cnt;if(ql<=l&&qr>=r)return (void)(vis[p]=1);
    ll mid=(l+r)>>1;
    if(ql<mid)build(ls[p],l,mid,ql,qr);
    if(qr>mid)build(rs[p],mid,r,ql,qr);
}
void query(int &p,ll l,ll r){
    if(!p)return (void)(p=++cnt,Pre[p]=suf[p]=(r-l)%P,ans=add(ans,calc((r-l)%P)));
    if(l+1==r)return (void)(suf[p]=Pre[p]=!vis[p],ans=add(ans,suf[p]));
    ll mid=(l+r)>>1;query(ls[p],l,mid),query(rs[p],mid,r);
    ans=add(ans,1ll*suf[ls[p]]*Pre[rs[p]]%P);
    if(vis[p]&&!vis[ls[p]]&&!vis[rs[p]])ans=dec(ans,1);
    if(!vis[p]&&(vis[ls[p]]||vis[rs[p]]))ans=add(ans,1);
    Pre[p]=add(Pre[ls[p]],!vis[p]);
    if(!vis[ls[p]])Pre[p]=add(Pre[p],dec(Pre[rs[p]],!vis[rs[p]]));
    suf[p]=add(suf[rs[p]],!vis[p]);
    if(!vis[rs[p]])suf[p]=add(suf[p],dec(suf[ls[p]],!vis[ls[p]]));
}
int main(){
//	freopen("testdata.in","r",stdin);
    n=read(),m=read();
    while(m--)l=read(),r=read(),build(rt,0,n,l-1,r);
    query(rt,0,n);printf("%d
",ans);return 0;
}