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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#517437#1281. Longest Common Subsequencewy2025TL 0ms20076kbC++141.4kb2024-08-13 11:28:522024-08-13 11:28:53

Judging History

你现在查看的是最新测评结果

  • [2024-08-13 11:28:53]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:20076kb
  • [2024-08-13 11:28:52]
  • 提交

answer

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e6+10;
int n,m,a[2][N],L[2][N],R[2][N],s[2][N],cl[2],cr[2];//0a,1b

struct T_A{
	int c[N*4];
	
	void init(){for(int i=1;i<=2*(n+m);i++) c[i]=0;}
	
	int lowbit(int x){return x&-x;}
	
	void update(int x,int v){x+=n+m;for(int i=x;i<=2*(n+m);i+=lowbit(i)) c[i]=max(c[i],v);}
	
	int ask(int x){
		x+=n+m;
		int ans=0;
		for(int i=x;i;i-=lowbit(i)) ans=max(ans,c[i]);
		return ans;
	}
	
}t[2];

void solve(){
	cin>>n>>m;
	for(int i=1;i<=n;i++) cin>>a[0][i];
	for(int i=1;i<=m;i++) cin>>a[1][i];
	n=min(n,m);
	for(int j=0;j<2;j++){
		cl[j]=cr[j]=0;
		R[j][0]=n+1;
		t[j].init();
		for(int i=1;i<=n;i++){
			if(a[j][i]==1) L[j][++cl[j]]=i;
			s[j][i]=s[j][i-1]+(a[j][i]==2);
		}
		for(int i=n;i;i--)
			if(a[j][i]==3) R[j][++cr[j]]=i;
	}
	int ans=max(s[0][n],s[1][n]);
	for(int i=0,j=min(cr[0],cr[1]);~j;j--){
		while(i<min(cl[0],cl[1])&&L[0][i+1]<R[0][j]&&L[1][i+1]<R[1][j])
			i++,
			t[0].update(s[0][R[0][i]-1]-s[1][R[1][i]-1],i-s[0][L[0][i]]),
			t[1].update(s[0][L[0][i]]-s[1][L[1][i]],i-s[1][L[1][i]]);
		ans=max(ans,j+max(s[0][R[0][j]-1]+t[0].ask(s[0][L[0][j]]-s[1][L[1][j]]),
						  s[1][R[1][j]-1]+t[1].ask(s[0][R[0][j]-1]-s[1][R[1][j]-1])));
	}
	cout<<ans<<"\n";
}

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0),cout.tie(0);
	int t;
	cin>>t;
	while(t--) solve();
	return 0;
}

Details

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Test #1:

score: 100
Accepted
time: 0ms
memory: 20076kb

input:

3
3 3
1 2 3
1 2 3
3 3
1 1 1
1 1 2
3 3
1 3 2
1 2 3

output:

3
2
2

result:

ok 3 tokens

Test #2:

score: -100
Time Limit Exceeded

input:

139889
1 10
2
2 1 2 2 3 1 1 2 3 3
7 1
3 2 3 3 1 1 2
2
6 1
2 1 3 1 1 1
1
8 7
3 1 3 3 2 2 3 1
3 2 2 1 3 3 3
10 3
3 2 1 1 2 2 1 1 1 1
3 1 1
5 2
1 2 1 3 1
1 2
1 4
1
3 3 3 3
7 2
3 1 2 1 2 2 3
3 2
6 2
3 1 1 2 1 3
1 3
1 4
1
3 1 1 3
4 2
2 3 2 3
1 3
1 8
1
1 1 3 1 3 3 3 1
4 1
1 3 3 3
1
10 10
3 1 2 1 2 2 2 2 1...

output:


result: