# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
329919 | ChrisGe123 | Palembang Bridges (APIO15_bridge) | C++14 | 11 ms | 620 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"
using namespace std;
typedef long long ll;
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <class T> using Tree = tree<T, null_type, less<T>,
rb_tree_tag, tree_order_statistics_node_update>;
//g++ -std=c++17 -O2 -fsanitize=undefined /usr/local/Cellar/gcc/10.2.0/include/c++/10.2.0/x86_64-apple-darwin19/bits/stdc++.h
#define maxn 100005
#define mod 1000000007
int n, k;
pair<ll, pair<ll, ll> > interval[maxn];
ll ans;
Tree<pair<int, int> > endpoints;
void setupIO(string s)
{
ios_base::sync_with_stdio(false);
cin.tie(0);
freopen((s + ".in").c_str(), "r", stdin);
freopen((s + ".out").c_str(), "w", stdout);
}
int main()
{
//setupIO("bridges");
ios_base::sync_with_stdio(false);
cin.tie(0);
cin >> k >> n;
int counter = 0;
ll pre = 0;
endpoints.clear();
for (int i=0; i<n; i++)
{
char p, q;
ll s, t;
cin >> p >> s >> q >> t;
if (p == q)
{
pre += abs(s - t);
}
else
{
interval[counter++] = make_pair(s+t, make_pair(min(s, t), max(s, t)));
}
}
n = counter;
if (k == 1)
{
ans = 0;
//we just want to find the median of the endpoints
for (int i=0; i<n; i++)
{
endpoints.insert(make_pair(interval[i].second.first, i));
endpoints.insert(make_pair(interval[i].second.second, i));
}
for (auto k: endpoints)
{
cerr << k.first << " ";
}
cerr << endl;
ll x = (endpoints.find_by_order(n-1)->first + endpoints.find_by_order(n)->first) >> 1;
for (int i=0; i<n; i++)
{
ll l = interval[i].second.first; ll r = interval[i].second.second;
if (l <= x && x <= r)
{
ans += r-l+1;
}
else if (x < l)
{
ans += 2 * (l -x) + r -l + 1;
}
else
{
ans += 2 * (x-r) + r - l + 1;
}
}
}
else
{
ans = LLONG_MAX;
sort(interval, interval+n); //sorted in order of increasing midpoint
for (int i=0; i<n-1; i++)
{
//things 0 to i take the first bridge, i+1 to n-1 take the second bridge
endpoints.clear();
for (int j=0; j<=i; j++)
{
endpoints.insert(make_pair(interval[j].second.first, j));
endpoints.insert(make_pair(interval[j].second.second, j));
}
ll x = (endpoints.find_by_order(i)->first + endpoints.find_by_order(i+1)->first) >> 1;
endpoints.clear();
for (int j=i+1; j<n; j++)
{
endpoints.insert(make_pair(interval[j].second.first, j));
endpoints.insert(make_pair(interval[j].second.second, j));
}
ll y = (endpoints.find_by_order(n-i-2)->first + endpoints.find_by_order(n-i-1)->first) >> 1;
//cerr << i << endl;
//cerr << x << " " << y << endl;
ll temp = 0;
for (int j=0; j<=i; j++)
{
ll l = interval[j].second.first; ll r = interval[j].second.second;
if (l <= x && x <= r)
{
temp += r-l+1;
}
else if (x < l)
{
temp += 2 * (l -x) + r -l + 1;
}
else
{
temp += 2 * (x-r) + r - l + 1;
}
}
for (int j=i+1; j<n; j++)
{
ll l = interval[j].second.first; ll r = interval[j].second.second;
if (l <= y && y <= r)
{
temp += r-l+1;
}
else if (y < l)
{
temp += 2 * (l - y) + r -l + 1;
}
else
{
temp += 2 * (y-r) + r - l + 1;
}
}
ans = min(ans, temp);
}
}
cout << ans + pre << endl;
}
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