Submission #306790

#	Time	Username	Problem	Language	Result	Execution time	Memory
306790		MarcoMeijer	Palembang Bridges (APIO15_bridge)	C++14	100 / 100	214 ms	15948 KiB

bridge

#include <bits/stdc++.h>
using namespace std;
 
// macros
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ll, ll> lll;
typedef tuple<int, int, int> iii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<ll> vll;
typedef vector<lll> vlll;
#define REP(a,b,c) for(int a=int(b); a<int(c); a++)
#define RE(a,c) REP(a,0,c)
#define RE1(a,c) REP(a,1,c+1)
#define REI(a,b,c) REP(a,b,c+1)
#define REV(a,b,c) for(int a=int(c-1); a>=int(b); a--)
#define FOR(a,b) for(auto& a : b)
#define all(a) a.begin(), a.end()
#define INF 1e18
#define EPS 1e-9
#define pb push_back
#define popb pop_back
#define fi first
#define se second
#define sz size()
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
 
// input
template<class T> void IN(T& x) {cin >> x;}
template<class H, class... T> void IN(H& h, T&... t) {IN(h); IN(t...); }
 
// output
template<class T1, class T2> void OUT(const pair<T1,T2>& x);
template<class T> void OUT(const vector<T>& x);
template<class T> void OUT(const T& x) {cout << x;}
template<class H, class... T> void OUT(const H& h, const T&... t) {OUT(h); OUT(t...); }
template<class T1, class T2> void OUT(const pair<T1,T2>& x) {OUT(x.fi,' ',x.se);}
template<class T> void OUT(const vector<T>& x) {RE(i,x.size()) OUT(i==0?"":" ",x[i]);}
template<class... T> void OUTL(const T&... t) {OUT(t..., "\n"); }
template<class H> void OUTLS(const H& h) {OUTL(h); }
template<class H, class... T> void OUTLS(const H& h, const T&... t) {OUT(h,' '); OUTLS(t...); }
 
//===================//
//  Added libraries  //
//===================//
 
//===================//
//end added libraries//
//===================//
 
void program();
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    program();
}
 
 
//===================//
//   begin program   //
//===================//
 
const int MX = 1e5+2;

ll k, n;
char p[MX], q[MX];
ll s[MX], t[MX];
ll ans;
map<ll,ll> mp;

priority_queue<ll                > maxHeap;
priority_queue<ll,vll,greater<ll>> minHeap;
ll smLeft, smRight;

void clearHeap() {
    while(!maxHeap.empty()) maxHeap.pop();
    while(!minHeap.empty()) minHeap.pop();
    smLeft = 0;
    smRight = 0;
}
void addPoint(ll x) {
    if(!maxHeap.empty() && x > maxHeap.top()) {
        minHeap.push(x);
        smRight += x;
    } else {
        maxHeap.push(x);
        smLeft += x;
    }
    while(maxHeap.size() > minHeap.size()) {
        ll x = maxHeap.top(); maxHeap.pop();
        smLeft -= x;
        smRight += x;
        minHeap.push(x);
    }
    while(maxHeap.size() < minHeap.size()) {
        ll x = minHeap.top(); minHeap.pop();
        smRight -= x;
        smLeft += x;
        maxHeap.push(x);
    }
}
ll getHeap() {
    ll x = maxHeap.top();
    return maxHeap.size()*x - smLeft + smRight - minHeap.size()*x;
}

void program() {
    IN(k,n);
    RE(i,n) IN(p[i],s[i],q[i],t[i]);
    RE(i,n) if(s[i] > t[i]) swap(s[i], t[i]);
    RE(i,n) s[i] *= 2, t[i] *= 2;

    RE(_,k) {
        clearHeap();
        priority_queue<lll> midPoints;
        RE(i,n) if(p[i] != q[i]) {
            if(_) {
                midPoints.push({-(t[i]+s[i])/2,i});
            } else {
                midPoints.push({ (t[i]+s[i])/2,i});
            }
        }
        ll lst = -1;
        while(!midPoints.empty()) {
            lll p = midPoints.top(); midPoints.pop();
            if(_) {
                p.fi = -p.fi;
                if(p.fi != lst && lst != -1)
                    mp[p.fi] += getHeap();
            }
            int i = p.se;
            addPoint(s[i]);
            addPoint(t[i]);
            if(!_)
                mp[p.fi] = (ans = getHeap());
            lst = p.fi;
        }
    }

    if(k == 2) {
        ans = INF;
        FOR(p,mp) ans = min(ans, p.se);
        if(mp.empty()) ans = 0;
    }
    RE(i,n) if(p[i] == q[i]) ans += abs(s[i]-t[i]);
    RE(i,n) if(p[i] != q[i]) ans += 2;
    ans /= 2ll;
    OUTL(ans);
}

• Subtask 1: 8 / 8
• Subtask 2: 14 / 14
• Subtask 3: 9 / 9
• Subtask 4: 32 / 32
• Subtask 5: 37 / 37