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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#142033#6733. Moniphant SleepOvO_ZuoWA 634ms50764kbC++142.4kb2023-08-18 11:27:092023-09-05 22:11:15

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你现在查看的是最新测评结果

  • [2023-09-05 22:11:15]
  • 管理员手动重测本题所有提交记录
  • 测评结果:WA
  • 用时:634ms
  • 内存:50764kb
  • [2023-08-18 11:27:11]
  • 评测
  • 测评结果:0
  • 用时:640ms
  • 内存:50740kb
  • [2023-08-18 11:27:09]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
const int N=5e5+5;
int n,q;
struct seg{
	int z,f,der,tz,tf;
	bool cl,mn;
	seg(int a=0,int b=0,int c=0,int d=0,int e=0,bool f=0,bool g=0):z(a),f(b),der(c),tz(d),tf(e),cl(f),mn(g){}
	void init()
	{
		z=f=der=tz=tf=0;
		cl=mn=0;
	}
	void add(int x){ der+=x;}
	void sub(int x)
	{
		der+=x;
		if(der<0)
		{
			mn=0;
			tz+=x;
			der=0;
		}		
		if(tz<0) tf+=tz,tz=0;
	}
	void upd()
	{
		if(mn) return;
		tz+=der;
		der=0;
		mn=1;
	}
	void clear()
	{
		if(!mn) tz+=der;
		mn=0;der=0;
		if(cl) return;
		if(z+tf<0) f+=z+tf,z=tz;
		else z+=tz+tf;
		tz=tf=0;
		cl=1;		
	}
	seg operator+(seg a)
	{
		seg res=seg(z,f,der,tz,tf,cl,mn);
		res.sub(a.f),res.add(a.z);
		if(a.cl) res.clear();
		res.sub(a.tf),res.add(a.tz);
		if(a.mn) res.upd();
		res.add(a.der);		
		//cout<<res.z<<" "<<res.f<<" "<<res.tz<<" "<<res.tf<<" "<<res.der<<" "<<res.cl<<" "<<res.mn<<endl;
		return res;
	}
	int v(){ return z+f+der+tz+tf;}
}shu[N<<2];
void build(int l,int r,int idx)
{
	if(l==r)
	{
		shu[idx].z=500000;
		return;
	}
	int mid=(l+r)>>1;
	build(l,mid,idx<<1),build(mid+1,r,idx<<1|1);
}
void push_down(int idx)
{
	shu[idx<<1]=shu[idx<<1]+shu[idx];
	shu[idx<<1|1]=shu[idx<<1|1]+shu[idx];
	shu[idx].init();
}
void modify(int l,int r,int tl,int tr,seg vv,int idx)
{
	if(l>=tl&&r<=tr)
	{
		shu[idx]=shu[idx]+vv;
		return;
	}
	push_down(idx);
	int mid=(l+r)>>1;
	if(mid>=tl) modify(l,mid,tl,tr,vv,idx<<1);
	if(mid<tr) modify(mid+1,r,tl,tr,vv,idx<<1|1);
}
int query(int l,int r,int tar,int idx)
{
	if(l==r) return shu[idx].v();
	push_down(idx);
	int mid=(l+r)>>1;
	if(mid>=tar) return query(l,mid,tar,idx<<1);
	else return query(mid+1,r,tar,idx<<1|1);
}
int main()
{
	scanf("%d%d",&n,&q);
	int i,op,l,r;
	build(1,n,1);
	for(i=1;i<=q;i++)
	{
		scanf("%d%d%d",&op,&l,&r);
		if(op==1) modify(1,n,l,r,seg(1,0,0,0,0,0,0),1);
		else if(op==2) modify(1,n,l,r,seg(0,-1,0,0,0,0,0),1);
		else if(op==3) modify(1,n,l,r,seg(0,0,0,0,0,0,1),1);
		else if(op==4) modify(1,n,l,r,seg(0,0,0,0,0,1,0),1);
		else printf("%d\n",query(1,n,l,1));
		//cout<<"query:"<<query(1,n,3,1)<<endl;
	}
	return 0;
}
/*
一/二操作为区间 +1/-1
四操作为跳转至有效的3操作中最小的后,将所有其余信息清空
不妨每次将清空后的位置视为0点,记录变化量
三操作时,只有更新最小值时需要操作:
	将最小值记为当前位置,并将变化量清空
*/

Details

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Test #1:

score: 100
Accepted
time: 4ms
memory: 50636kb

input:

1 9
1 1 1
1 1 1
1 1 1
3 1 1
2 1 1
1 1 1
1 1 1
4 1 1
5 1 1

output:

500004

result:

ok 1 number(s): "500004"

Test #2:

score: 0
Accepted
time: 7ms
memory: 50692kb

input:

3 7
2 1 3
3 1 3
1 1 3
1 1 3
5 1 1
4 1 3
5 1 1

output:

500001
499999

result:

ok 2 number(s): "500001 499999"

Test #3:

score: -100
Wrong Answer
time: 634ms
memory: 50764kb

input:

500000 500000
2 132991 371170
5 15281 15281
1 278642 397098
2 152103 176573
2 13797 47775
3 139320 370045
3 79054 432340
3 82556 212775
4 270171 469418
5 148000 148000
3 371255 401414
5 71051 71051
2 358945 473071
2 231663 265401
2 20243 58131
1 247710 313373
5 154549 154549
1 17317 233265
5 37602 3...

output:

500000
499999
500000
499998
499999
500000
500000
499997
500000
499997
499997
499999
499997
500000
499997
500000
499999
500001
499999
499999
499997
499997
499996
499998
500000
500001
500001
499996
499998
499999
499996
499999
500001
499998
500000
500002
500002
499997
500001
499997
499995
499998
500004...

result:

wrong answer 13th numbers differ - expected: '499998', found: '499997'