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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#417562#1833. DeletingHarry27182RE 1ms5688kbC++141.9kb2024-05-22 19:43:242024-05-22 19:43:24

Judging History

你现在查看的是最新测评结果

  • [2024-05-22 19:43:24]
  • 评测
  • 测评结果:RE
  • 用时:1ms
  • 内存:5688kb
  • [2024-05-22 19:43:24]
  • 提交

answer

#include<bits/stdc++.h>
#define ull unsigned long long
using namespace std;
int n,ok[4005][4005],x;pair<int,int>p[4000005];ull num[64];
struct Bitset
{
	ull val[75];
	int get(int x){return (val[x/64]>>(x&63))&1;}
	void upd(int x){val[x/64]^=(1llu<<(x&63));}
	int _Find_first()
	{
		for(int i=0;i<=n/64;i++)
		{
			if(!val[i])continue;
			return __builtin_ctzll(val[i])+i*64;
		}
		return n+1;
	}
	int _Find_next(int x)
	{
		int p=x/64,q=x&63;
		if(val[p]&num[q])return __builtin_ctzll(val[p]&num[q])+p*64;
		for(int i=p+1;i<=n/64;i++)
		{
			if(!val[i])continue;
			return __builtin_ctzll(val[i])+i*64;
		}
		return n+1;
	}
}L[4005],R[4005],nL[4005],nR[4005];
Bitset operator &(Bitset x,Bitset y)
{
	Bitset res;
	for(int i=0;i<=n/64;i++)res.val[i]=x.val[i]&y.val[i];
	return res;
}
void insert(int l,int r)
{
	assert(L[l].get(r));
	if(l-1>=1&&r+1<=n&&ok[l-1][r+1]&&!L[l-1].get(r+1))
	{
		L[l-1].upd(r+1);R[r+1].upd(l-1);nL[l-1].upd(r+1);nR[r+1].upd(l-1);
		insert(l-1,r+1);
	}
	Bitset now=R[l-1]&nR[r];
	for(int i=now._Find_first();i<=n;i=now._Find_next(i))
	{
		L[i].upd(r);R[r].upd(i);nL[i].upd(r);nR[r].upd(i);
		insert(i,r);
	}
	now=L[r+1]&nL[l];
	for(int i=now._Find_first();i<=n;i=now._Find_next(i))
	{
		L[l].upd(i);R[i].upd(l);nL[l].upd(i);nR[i].upd(l);
		insert(l,i);
	}
}
int solve(int l,int r)
{
	ok[l][r]=1;
	if((l+1==r||L[l+1].get(r-1))&&(!L[l].get(r)))
	{
		L[l].upd(r);R[r].upd(l);nL[l].upd(r);nR[r].upd(l);
		insert(l,r);
	}
	return L[1].get(n);
}
int main()
{
	cin.tie(0)->sync_with_stdio(0);
	cin>>n;
	for(int i=0;i<64;i++)
	{
		for(int j=i+1;j<64;j++)num[i]|=(1llu<<j);
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)nL[i].upd(j),nR[i].upd(j);
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=i+1;j<=n;j+=2)cin>>x,p[x]={i,j};
	}
	for(int i=1;i<=(n/2)*(n/2);i++)
	{
		if(solve(p[i].first,p[i].second)){cout<<i;return 0;}
	}
	return 0;
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 5540kb

input:

2
1

output:

1

result:

ok 1 number(s): "1"

Test #2:

score: 0
Accepted
time: 0ms
memory: 5688kb

input:

6
2 1 3
4 5
6 7
8
9

output:

6

result:

ok 1 number(s): "6"

Test #3:

score: -100
Runtime Error

input:

10
20 21 2 11 25
3 24 18 8
6 17 7 5
22 4 23
14 15 1
19 16
12 10
13
9

output:


result: