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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#287449 | #5407. 基础图论练习题 | JJ | 0 | 44ms | 16256kb | C++20 | 4.6kb | 2023-12-20 16:07:35 | 2023-12-20 16:07:36 |
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<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;
}
Details
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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%