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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 |
---|---|---|---|---|---|---|---|---|---|
#287468 | #5407. 基础图论练习题 | JJ | 0 | 38ms | 18976kb | C++20 | 4.9kb | 2023-12-20 16:52:01 | 2023-12-20 16:52:02 |
Judging History
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<to[u].size();j++)
{
int v=pos[to[u][j]];
if(b[v]!=ps) continue;
if(v==(i+1)%m) continue;
//cout<<v<<"~"<<i<<endl;
if(v<i) 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;i++) cout<<tv[i]<<" ";puts("");
//cout<<prf[m-1]<<" "<<prf[0]<<":"<<tot+v0[m-2]<<endl;
}
for(int i=0;i<m-1;i++)
{
int u=prf[i];
for(int j=0;j<to[u].size();j++)
{
int v=pos[to[u][j]];
if(b[v]!=ps) continue;
if(v==(i+1)%m) continue;
if(v>i) val[v]++,val[i+m]--;
else val[v+m]++,val[i+m]--;
}
for(int j=0;j<fr[u].size();j++)
{
int v=pos[fr[u][j]];
if(b[v]!=ps) continue;
if(i==(v+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 j=1;j<=m+i;j++) tv[j]=tv[j-1]+val[j];
for(int j=1;j<=m+i;j++) v0[j]=v0[j-1]+(tv[j]==0);
//for(int j=0;j<=m+i;j++) cout<<tv[j]<<" ";puts("");
ans=(ans+1ll*gval(prf[i],prf[i+1])*(tot+v0[m+i-1]-v0[i])%mod)%mod;
//cout<<prf[i]<<" "<<prf[i+1]<<":"<<tot+v0[m+i-1]-v0[i]<<endl;
}
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: 38ms
memory: 18976kb
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:
44015257 13 352453823 428637146 182 0 1983 2 302319119 273158175 891998853 49489 2 801640988 115 3219583 2527 0 471087262 560413560 2902272 40095 39489 3769610 947076164 780131398 220081587 721573443 374182802 65360 269750275 227868047 53520 2162943 119367680 405634000 162 168 235 5245951 268582975 ...
result:
wrong answer 1st numbers differ - expected: '281603732', found: '44015257'
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%