bzoj4358: permu

莫队算法，用线段树维护最长连续1，复杂度O(nsqrt(m)logn)

刚开始TLE了，看了claris大爷的blog说是kd-tree，然而并不会kd-tree……

然后就打算弃疗了...弃疗之前加了点常数优化，然后就AC了...（滑稽

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define N 50004
#define ls(a) (a<<1)
#define rs(a) (a<<1^1)

using namespace std;
inline int read(){
	int ret=0;char ch=getchar();
	while (ch<'0' || ch>'9') ch=getchar();
	while ('0'<=ch && ch<='9'){
		ret=ret*10-48+ch;
		ch=getchar();
	}
	return ret;
}

struct query{
	int x,y,key,id;
} q[N];
inline bool operator <(const query &a,const query &b){
	return (a.key<b.key||a.key==b.key&&((a.y<b.y)^(a.key&1)));
}

struct STnode{
	int len;
	int maxs,maxl,maxr;
} t[N*4];
int pos[N];
inline void PushUp(int x){
	t[x].maxs=max(t[ls(x)].maxr+t[rs(x)].maxl,max(t[ls(x)].maxs,t[rs(x)].maxs));
	t[x].maxl=t[ls(x)].maxl+(t[ls(x)].maxl==t[ls(x)].len)*t[rs(x)].maxl;
	t[x].maxr=t[rs(x)].maxr+(t[rs(x)].maxr==t[rs(x)].len)*t[ls(x)].maxr;
}
void build(int x,int l,int r){
	t[x].maxs=t[x].maxl=t[x].maxr=0;
	t[x].len=r-l+1;
	if (t[x].len==1){pos[l]=x;return;}
	int mid=(l+r)/2;
	build(x<<1,l,mid);
	build(x<<1^1,mid+1,r);
}
void update(int p){
	int x=pos[p];
	t[x].maxl=t[x].maxr=(t[x].maxs^=1);
	while (x!=1) PushUp(x=x>>1);
}
int getans(){
	return t[1].maxs;
}

int a[N],n;
int l,r;
void move(int L,int R){
	for (;l>L;--l) update(a[l-1]);
	for (;r<R;++r) update(a[r+1]);
	for (;l<L;++l) update(a[l]);
	for (;r>R;--r) update(a[r]);
}

int ans[N];

int main(){
	n=read();int Q=read();
	for (int i=1;i<=n;++i) a[i]=read();
	for (int i=1;i<=Q;++i){
		q[i].x=read();q[i].y=read();
		q[i].key=sqrt(q[i].x);
		q[i].id=i;
	}
	sort(q+1,q+Q+1);
	l=1;r=0;
	build(1,1,n);
	for (int i=1;i<=Q;++i){
		if (q[i].x>q[i].y){ans[q[i].id]=0;continue;}
		move(q[i].x,q[i].y);
		ans[q[i].id]=getans();
	}
	for (int i=1;i<=Q;++i) printf("%d\n",ans[i]);
	return 0;
}