이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#include "keys.h"
using namespace std;
const int maxn = 3e5 + 5;
int n,m,p[maxn],r[maxn];
// DSU
int par[maxn]; // the root of the group
int sz[maxn]; // the size of the group
int find(int x)
{
if (par[x]==x) return x;
return par[x] = find(par[x]);
}
bool have[maxn],vis[maxn],mk[maxn];
int hd,tl,q[maxn];
vector<pair<int,int>> g[maxn];
vector<int> vec[maxn];
vector<int> ans;
int bfs(int s,bool flag=0)
{
hd = 1;
tl = 0;
int ret = 0;
q[++tl] = s;
vis[s] = 1;
while (hd<=tl)
{
int x = q[hd++];
if (!have[r[x]])
{
while (!vec[r[x]].empty())
{
int y=vec[r[x]].back();
if (find(y)!=s) ret = find(y);
if (!vis[y]) q[++tl] = y, vis[y] = 1;
vec[r[x]].pop_back();
}
have[r[x]]=1;
}
for (pair<int,int> A : g[x])
{
int y=A.first;
if (have[A.second])
{
if (find(y)!=s) ret = find(y);
if (!vis[y]) q[++tl] = y, vis[y] = 1;
}
else vec[A.second].push_back(y);
}
if (ret) break;
}
for (int i=1;i<=tl;i++) vis[q[i]] = 0;
for (int i=1;i<hd;i++)
{
int x = q[i];
have[r[x]] = 0;
for (pair<int,int> A : g[x])
vec[A.second].clear();
}
if (flag)
{
for (int i=1;i<=tl;i++)
ans[q[i]-1] = tl;
}
return ret;
}
vector<int> find_reachable(vector<int> R, vector<int> U, vector<int> V, vector<int> C)
{
n = R.size(); m = U.size();
for (int i=0;i<n;i++) r[i+1] = R[i] + 1;
ans = vector<int> (n,n+1);
for (int i=0;i<m;i++)
{
int x,y,z;
x = U[i]+1;
y = V[i]+1;
z = C[i]+1;
g[x].push_back({y,z}),g[y].push_back({x,z});
}
// initialize the DSU
for (int i=1;i<=n;i++)
{
par[i] = i;
sz[i] = 1;
}
while (true)
{
bool flag = false;
for (int i=1;i<=n;i++) mk[i] = false;
for (int i=1;i<=n;i++)
{
if(find(i)==i && !mk[i])
{
int t = bfs(i);
if(t)
{
par[i] = t;
mk[t] = true;
flag = false;
}
else mk[i] = 1;
}
}
if (!flag) break;
}
// bfs only the roots
for (int i=1;i<=n;i++)
{
if (find(i)==i) bfs(i,1);
}
// formatting the answer: ans[i] is the number of rooms reachable
int mn = n+1;
for (int i=1;i<=n;i++) mn = min(mn,ans[i-1]);
for (int i=1;i<=n;i++)
{
if (mn==ans[i-1]) ans[i-1] = 1;
else ans[i-1] = 0;
}
return ans;
}
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