# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
81601 | ToadDaveski | Evacuation plan (IZhO18_plan) | C++11 | 446 ms | 42032 KiB |
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>
#define ll long long
#define fr first
#define sc second
using namespace std;
vector <pair <ll,ll> > v[100001];
ll dp[100001],us[100001],pr[100001];
unordered_set <ll> an[100001];
ll ans[100001];
priority_queue <pair <ll,ll> > q;
vector <pair <ll,pair <ll,ll> > > edges;
ll dfs(ll x,ll y)
{
us[x]=-1;
if (x==y) return us[x]=dp[y];
ll ans=0;
for(auto to : v[x])
{
if (us[to.fr]==0)
dfs(to.fr,y);
if (us[to.fr]>0)
ans=max(ans,us[to.fr]);
}
return us[x]=min(ans,dp[x]);
}
bool cmp (pair <ll,ll> a, pair <ll,ll> b)
{
if (a.sc>=b.sc)
return 1;
return 0;
}
ll f(ll x)
{
if (pr[x]==x) return x;
else return pr[x]=f(pr[x]);
}
int main()
{
//ios_base::sync_with_stdio(0);
ll n,m,Q,k,i,j;
cin>>n>>m;
//cout<<3<<endl;
for(i=1;i<=n;i++)
dp[i]=1e9;
for(i=1;i<=m;i++)
{
ll x,y,z;
cin>>x>>y>>z;
v[x].push_back({y,z});
v[y].push_back({x,z});
}
cin>>k;
for(i=1;i<=k;i++)
{
ll x,y;
cin>>x;
q.push({0,x});
dp[x]=0;
}
//cout<<3<<endl;
while(!q.empty())
{
ll from=q.top().sc;
ll weight=-q.top().fr;
q.pop();
if (dp[from]<weight)
continue;
for(auto to : v[from])
{
if (dp[to.fr]>dp[from]+to.sc)
{
dp[to.fr]=dp[from]+to.sc;
q.push({-(dp[from]+to.sc),to.fr});
}
}
}
//cout<<3<<endl;
cin>>Q;
for(i=1;i<=Q;i++)
{
ll x,y;
scanf("%d%d", &x, &y);
an[x].insert(i);
an[y].insert(i);
}
for(i=1;i<=n;i++)
{
pr[i]=i;
for(auto to : v[i])
{
edges.push_back({min(dp[to.fr],dp[i]),{i,to.fr}});
}
}
//cout<<edges.size()<<endl;
sort(edges.begin(),edges.end());
//cout<<3<<endl;
for(i=edges.size()-1;i>=0;i--)
{
//cout<<i<<endl;
ll a=f(edges[i].sc.fr);
ll b=f(edges[i].sc.sc);
if (a==b) continue;
if (an[a].size()<an[b].size())
swap(a,b);
for(const int& tod : an[b])
{
if (an[a].count(tod))
{
ans[tod]=edges[i].fr;
an[a].erase(tod);
an[b].erase(tod);
}
else
{
an[a].insert(tod);
}
}
pr[b]=a;
}
for(i=1;i<=Q;i++)
printf("%d\n", ans[i]);
return 0;
}
/*
9 12
1 9 4
1 2 5
2 3 7
2 4 3
4 3 6
3 6 4
8 7 10
6 7 5
5 8 1
9 5 7
5 4 12
6 8 2
2
4 7
5
1 6
5 3
4 8
5 8
1 5
*/
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