QOJ.ac

QOJ

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
#287449#5407. 基础图论练习题JJ0 44ms16256kbC++204.6kb2023-12-20 16:07:352023-12-20 16:07:36

Judging History

你现在查看的是最新测评结果

  • [2023-12-20 16:07:36]
  • 评测
  • 测评结果:0
  • 用时:44ms
  • 内存:16256kb
  • [2023-12-20 16:07:35]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
int c[5005][5005],du[5005];
bool cmp(int x,int y){return du[x]<du[y];}
vector<int>fr[5005],to[5005];
void add(int aa,int bb)
{
	//cout<<aa<<"->"<<bb<<endl;
	fr[bb].push_back(aa);
	to[aa].push_back(bb);
	du[aa]++;
}
void init(int n)
{
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++) c[i][j]=0;
	for(int i=1;i<=n;i++) du[i]=0,fr[i].clear(),to[i].clear();
}
int vis[5005];
vector<int>solve(vector<int> vec)
{
	if(vec.size()==1) return vec;
	//for(int i=0;i<vec.size();i++)
	//cout<<vec[i]<<" ";puts("");
	vector<int>prf;int x=vec.back();vec.pop_back();
	for(int i=0;i<vec.size();i++) vis[vec[i]]=1;
	for(int i=0;i<fr[x].size();i++)
	if(vis[fr[x][i]]) du[fr[x][i]]--;
	for(int i=0;i<vec.size();i++) vis[vec[i]]=0;
	//删掉一个点在其它点的导出子图求强连通分量 
	sort(vec.begin(),vec.end(),cmp);
	long long sum=0;int las=-1;
	for(int i=0;i<vec.size();i++)
	{
		int u=vec[i];sum+=du[u]-las-1;
		//cout<<u<<":"<<du[u]<<endl;
		if(sum==(i-las)*(i-las-1)/2)
		{
			//cout<<u<<endl;
			vector<int>now,tmp;
			for(int j=las+1;j<=i;j++)
			du[vec[j]]-=las+1,now.push_back(vec[j]);
			tmp=solve(now);las=i;sum=0;
			for(int j=tmp.size()-1;j>=0;j--)
			prf.push_back(tmp[j]);
			//cout<<u<<"is fine"<<endl;
		}
	}
	for(int i=0;i<prf.size()/2;i++) swap(prf[i],prf[prf.size()-1-i]);
	//for(int i=0;i<prf.size();i++) cout<<prf[i]<<" ";puts("");
	prf.push_back(x);
	return prf;
}
char s[5005];
int md[3000005],b[5005],M=3000000,ans,tot;
int tv[10005],val[10005],v0[10005],pos[5005],si[5005];
int gval(int i,int j){if(i<j)swap(i,j);return md[(i-2)*(i-1)/2+j-1];}
void getans(vector<int>vec,int ps)
{
	vector<int>prf=solve(vec);int m=vec.size();
	//for(int i=0;i<m;i++) cout<<prf[i]<<" ";puts("");
	//吾有三贡献:哈密顿路径上、哈密顿路径外、到出度更小的连通块 
	for(int i=0;i<m;i++) pos[prf[i]]=i;
	for(int i=0;i<m*2;i++) val[i]=0;
	for(int i=0;i<m;i++)
	{
		int u=prf[i];
		for(int j=0;j<fr[u].size();j++)
		{
			int v=pos[fr[u][j]];
			if(i==(v+1)%m) continue;
			val[v]++;val[i]--;
		}
	}
	if(m!=1)
	{
		tv[0]=val[0];v0[0]=(tv[0]==0);
		for(int i=1;i<m;i++) tv[i]=tv[i-1]+val[i];
		for(int i=1;i<m;i++) v0[i]=v0[i-1]+(tv[i]==0);
		ans=(ans+1ll*gval(prf[m-1],prf[0])*(tot+v0[m-1])%mod)%mod;
	}
	for(int i=0;i<m-1;i++)
	{
		int u=prf[i];
		for(int j=0;j<fr[u].size();j++)
		{
			int v=pos[fr[u][j]];
			if(i==(v+1)%m) continue;
			if(v>i) val[v]++,val[i+m]--;
			else val[v+m]++,val[i+m]--;
		}
		for(int j=0;j<to[u].size();j++)
		{
			int v=pos[to[u][j]];
			if(v==(i+1)%m) continue;
			if(v>i) val[i]--,val[v]++;
			else val[i]--,val[v+m]++;
		}//这里理论上从i位开始前缀和就是对的 
		tv[0]=val[0];v0[0]=(tv[0]==0);
		for(int i=1;i<m;i++) tv[i]=tv[i-1]+val[i];
		for(int i=1;i<m;i++) v0[i]=v0[i-1]+(tv[i]==0);
		ans=(ans+1ll*gval(prf[i],prf[i+1])*(tot+v0[m+i-1]-v0[i])%mod)%mod;
	}
	for(int ii=0;ii<m;ii++)
	{
		int i=vec[ii];//cout<<i<<endl;
		for(int j=0;j<to[i].size();j++)
		{
			int va=gval(to[i][j],i);
			//cout<<to[i][j]<<" "<<i<<":"<<va<<endl;
			if(b[to[i][j]]==ps){
				if(pos[to[i][j]]!=(pos[i]+1)%m)
				ans=(ans+1ll*va*tot%mod)%mod;
			}
			else
			{
				if(b[to[i][j]]==ps-1&&si[ps]==1&&si[ps-1]==1)
				ans=(ans+1ll*va*tot%mod)%mod;
				//仍然是链,形态会变成 1->v->u->v+1->n 
				else ans=(ans+1ll*va*(tot-(ps-b[to[i][j]]))%mod)%mod;
			}
		}
	}
}
int main()
{
	int tim;cin>>tim;md[0]=1;
	for(int i=1;i<=M;i++) md[i]=2ll*md[i-1]%mod;
	while(tim--)
	{
		int n;cin>>n;ans=tot=0;init(n);
		for(int i=2;i<=n;i++)
		{
			scanf("%s",s+1);
			for(int j=1;j<=(i+3)/4;j++)
			{
				int val=(s[j]<='9'?s[j]-'0':s[j]-'A'+10);
				for(int k=1;k<=4;k++)
				c[i][(j-1)*4+k]=(val>>k-1)&1;
			}
			for(int j=1;j<i;j++)
			if(c[i][j]) add(i,j);
			else add(j,i);
		}
		vector<int>vec;
		for(int i=1;i<=n;i++) vec.push_back(i);
		sort(vec.begin(),vec.end(),cmp);
		long long sum=0;int las=-1;
		for(int i=0;i<vec.size();i++)
		{
			int uu=vec[i];sum+=du[uu]-las-1;
			//cout<<u<<":"<<du[u]<<endl;
			if(sum==(i-las)*(i-las-1)/2)
			{
				si[++tot]=0;
				for(int j=las+1;j<=i;j++)
				b[vec[j]]=tot,si[tot]++;
				las=i;sum=0;
			}
		}
		sum=0;las=-1;int tmp=0;
		for(int i=0;i<vec.size();i++)
		{
			int uu=vec[i];sum+=du[uu]-las-1;
			//cout<<u<<":"<<du[u]<<endl;
			if(sum==(i-las)*(i-las-1)/2)
			{
				vector<int>now;
				for(int j=las+1;j<=i;j++)
				du[vec[j]]-=las+1,now.push_back(vec[j]);
				getans(now,++tmp);las=i;sum=0;
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}

詳細信息

Subtask #1:

score: 0
Wrong Answer

Test #1:

score: 0
Wrong Answer
time: 44ms
memory: 16256kb

input:

10000
100
1
2
2
8
C0
F0
27
78
AE1
C01
511
D87
EF20
3873
2742
73D0
DC9B0
FB2A3
9C011
9B4E0
95DC00
A7B980
F43531
6A6245
5347BE0
1A6C8A1
88E46D6
64CF3AE
D25F63C1
C894E4C3
1C0AFD73
EC1C3F9A
087CE17C0
22149A380
B28038AF1
B9CA21C7F
D78F5307C1
49045489A2
72C4DE6FD1
7713F40D05
EEE8878EEC1
310E62812B1
DA9D5B...

output:

819140252
13
335611071
411859922
97
0
1983
2
302121487
270536735
302064061
33088
2
801655836
61
3219455
2521
0
470988950
560397240
2894079
37920
36864
3769598
947010596
780133438
40212386
721049155
194313737
64424
236715889
211089805
41215
1990669
790845987
381345265
99
111
163
2098175
268582943
13
...

result:

wrong answer 1st numbers differ - expected: '281603732', found: '819140252'

Subtask #2:

score: 0
Skipped

Dependency #1:

0%

Subtask #3:

score: 0
Skipped

Dependency #1:

0%

Subtask #4:

score: 0
Skipped

Dependency #1:

0%

Subtask #5:

score: 0
Skipped

Dependency #1:

0%