线段树

20+30+68(2/6) -> 100+30+68(1/6)
T1线段树+二分板子，由于还有很多人也改出来了就把代码放了吧

#include
using namespace std;
long long t[400050],a[100005],lazy[400050],n,m,b,c,d,e,ll,rr,al,ar,mid;
long long lowbit(long long x){return x&-x;}
void add(long long k,long long l,long long r,long long v){
    lazy[k]=v;
    t[k]=v*(r-l+1);
}
void pd(long long k,long long l,long long r,long long m){
    if(lazy[k]==-1){return;}
    if(l==r){return;}
    add(k*2,l,m,lazy[k]);
    add(k*2+1,m+1,r,lazy[k]);
    lazy[k]=-1;
}
long long sum(long long k,long long l,long long r,long long x,long long y){
    if(x=r){
        return t[k];
    }
    long long m=(l+r)/2,ans=0;
    pd(k,l,r,m);
    if(xm) ans+=sum(k*2+1,m+1,r,x,y);
    return ans;
}
void cge(long long k,long long l,long long r,long long x,long long y,long long v){
    if(x=r){add(k,l,r,v);return;}
    long long m=(l+r)/2;
    pd(k,l,r,m);
    if(xm) cge(k*2+1,m+1,r,x,y,v);
    t[k]=t[k*2]+t[k*2+1];
}
int main(){
    freopen("explore.in","r",stdin);
    freopen("explore.out","w",stdout);
    cin>>n>>m;
    memset(lazy,0,sizeof(lazy));
    for(int i=1;i>b;
        if(b==1){
            cin>>c>>d;
            cge(1,1,n,c,d,1);
        }else if(b==2){
            cin>>c>>d;
            cge(1,1,n,c,d,0);
        }else{
            cin>>c;
            if(sum(1,1,n,1,c)==0){cout<<"INF\n";continue;}
            if(sum(1,1,n,c,n)==0){cout<0){cout<<"0\n";continue;}
            ll=1;rr=c;
            while(ll<rr){
                mid=(ll+rr+1)/2;
                if(sum(1,1,n,mid,c)==0){
                    al=mid;rr=mid-1;
                }else{ll=mid;}
            }
            ll=c;rr=n;
            while(ll<rr){
                mid=(ll+rr)/2;
                if(sum(1,1,n,c,mid)==0){
                    ar=mid;ll=mid+1;
                }else{rr=mid;}
            }
            //cout<<ar<<' '<<al<<endl;
            cout<<ar-al+1<<endl;
        }
    }
    return 0;
}

LCA和倍增

19+44+0(1/2)
是的我们只有2个人交了(我和wwh)
听不懂
下班