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 |
---|---|---|---|---|---|---|---|---|---|
#287485 | #5407. 基础图论练习题 | JJ | 0 | 38ms | 16372kb | C++20 | 5.0kb | 2023-12-20 17:20:28 | 2023-12-20 17:20:28 |
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[to[u][j]]!=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-2])%mod)%mod;
//for(int i=0;i<m;i++) cout<<tv[i]<<" ";puts("");
//cout<<prf[m-1]<<" "<<prf[0]<<":"<<tot+v0[m-2]<<" "<<ans<<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[to[u][j]]!=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[fr[u][j]]!=ps) continue;
if(i==(v+1)%m) continue;
//cout<<u<<"->"<<fr[u][j]<<":"<<v<<"~"<<i<<endl;
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]<<" "<<ans<<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;
}
详细
Subtask #1:
score: 0
Wrong Answer
Test #1:
score: 0
Wrong Answer
time: 38ms
memory: 16372kb
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:
618229169 13 352322751 395082706 84 0 1983 2 302055951 268439583 712129788 33103 2 801607196 123 3154047 2523 0 470956174 560380792 2883839 37023 38911 3245312 946945060 780123198 860343336 720524867 14444672 65224 269750273 194313613 37119 2171135 757291555 380296689 120 84 219 5245439 268582911 13...
result:
wrong answer 1st numbers differ - expected: '281603732', found: '618229169'
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%