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;
#define fi first
#define se second
#define eb emplace_back
#define mt make_tuple
#define all(x) (x).begin(), (x).end()
#define sz(x) int(x.size())
#define MOD 1000000007
typedef long long ll;
typedef pair <int, int> ii;
typedef pair <ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef long double ld;
const int mxn = 1e5+5;
bool DEBUG=0;
ll pref[mxn],suff[mxn];
bool cmp(const pll &a, const pll &b){
if(a.fi+a.se!=b.fi+b.se)return a.fi+a.se<b.fi+b.se;
else return a.fi<b.fi;
}
int main()
{
ios_base::sync_with_stdio(0); cin.tie(0);
ll k,n;
vll coor;
vector<pll>str;
cin >> k >> n;
ll ans = 0ll;
for(int i=0; i<n; i++){
char p,q; ll s,t;
cin >> p >> s >> q >> t;
if(p==q){
ans += abs(s-t);
}else{
if(s>t)swap(s,t);
ans+=1ll;
str.eb(s,t);
}
}
n=sz(str);
if(n==0){
//no bridge is needed
cout << ans << endl;
return 0;
}
if(k==1){
//k=1
for(ii x:str){
coor.eb(x.fi);
coor.eb(x.se);
}
sort(all(coor));
ll bridge1=coor[n-1],bridge2=coor[n];
ll res1=0ll,res2=0ll;
for(int i=0; i<n; i++){
if(bridge1<str[i].fi){
res1 += 2ll*(str[i].fi-bridge1);
}else if(bridge1>str[i].se){
res1 += 2ll*(bridge1-str[i].se);
}
if(bridge2<str[i].fi){
res2 += 2ll*(str[i].fi-bridge2);
}else if(bridge2>str[i].se){
res2 += 2ll*(bridge2-str[i].se);
}
}
ans += min(res1,res2);
}else{
//k=2
sort(all(str),cmp);
multiset<ll>sm,bg;
ll sum_sm=0ll,sum_bg=0ll;
for(int i=0; i<n-1; i++){ //bridge 1, pref
sm.insert(str[i].fi);
bg.insert(str[i].se);
sum_sm += str[i].fi;
sum_bg += str[i].se;
while(*sm.rbegin()>*bg.begin()){ // swap em, so the median is still at sm(last),bg(first)
ll dif = *sm.rbegin()-*bg.begin();
sum_sm -= dif;
sum_bg += dif;
sm.insert(*bg.begin());
bg.insert(*sm.rbegin());
sm.erase(--sm.end());
bg.erase(bg.begin());
}
//always throw to bg group, so easier to calc the dif between groups
pref[i] = sum_bg-sum_sm;
}
sm=bg=multiset<ll>();
sum_sm=sum_bg=0ll;
for(int i=n-1; i>0; i--){ //bridge 2, suff
sm.insert(str[i].fi);
bg.insert(str[i].se);
sum_sm += str[i].fi;
sum_bg += str[i].se;
while(*sm.rbegin()>*bg.begin()){ // swap em
ll dif = *sm.rbegin()-*bg.begin();
sum_sm -= dif;
sum_bg += dif;
sm.insert(*bg.begin());
bg.insert(*sm.rbegin());
sm.erase(--sm.end());
bg.erase(bg.begin());
}
suff[i] = sum_bg-sum_sm;
}
// combining pref_i+suff_(i+1) as pref and suff go to bridge 1 and 2 respectively
ll res = 1e18;
for(int i=0; i<n-1; i++){
res = min(res,pref[i]+suff[i+1]);
}
ans += res;
}
cout << ans << '\n';
return 0;
}
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