제출 #554811

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
554811		kwongweng	Palembang Bridges (APIO15_bridge)	C++17	0 / 100	1 ms	468 KiB

bridge

/*
Solution for APIO 2015 - Palembang Bridges
Tags :
*/
 
#include <bits/stdc++.h>
using namespace std;
 
#pragma GCC target ("avx2")
#pragma GCC optimization ("Ofast")
#pragma GCC optimization ("unroll-loops")
 
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef long double ld;
typedef pair<ll, ll> pll;
#define FOR(i, a, b) for(int i = a; i < b; i++)
#define ROF(i, a, b) for(int i = a; i >= b; i--)
#define ms memset
#define pb push_back
#define fi first
#define se second
 
ll MOD = 1000000007;
 
ll power(ll base, ll n){
	if (n == 0) return 1;
	if (n == 1) return base;
	ll halfn = power(base, n/2);
	if (n % 2 == 0) return (halfn * halfn) % MOD;
	return (((halfn * halfn) % MOD) * base) % MOD;
}
 
ll inverse(ll n){
	return power(n, MOD-2);
}
 
ll add(ll a, ll b){
	return (a+b) % MOD;
}
 
ll mul(ll a, ll b){
	a %= MOD;
	return (a*b) % MOD;
}
 
ll gcd(ll a, ll b){
    if (a == 1) return 1;
    if (a == 0) return b;
    return gcd(b%a, a);
}
 
void solve(){
    int k, n; cin >> k >> n;
    ll ans = 0;
    vector<pair<ll, ll>> loc;
    FOR(i,0,n){
        char z1, z2;
        ll l, r;
        cin >> z1 >> l >> z2 >> r;
        if (l > r) swap(l,r);
        if (z1 == z2){
            ans += (r-l);
        }else{
            ans += (r-l+1);
            loc.pb({l, r});
        }
    }
    n = loc.size();
    sort(loc.begin(), loc.end());
    vector<ll> l(n), r(n);
    FOR(i,0,n){
        l[i] = loc[i].fi;
        r[i] = loc[i].se;
    }
    if (k==1){
        sort(l.begin(), l.end());
        sort(r.begin(), r.end());
        ll mini = MOD*MOD;
        vector<ll> pre_r(n), suf_l(n);
        pre_r[0] = r[0];
        FOR(i,1,n){
            pre_r[i] = pre_r[i-1] + r[i];
        }
        suf_l[n-1] = l[n-1];
        ROF(i,n-2,0){
            suf_l[i] = suf_l[i+1] + l[i];
        }
        FOR(i,0,n){
            ll ans_r = r[i] * (ll) (i+1) - pre_r[i];
            if (r[i] >= l[n-1]){
                mini = min(mini, ans_r);
                continue;
            }
            ll pos_r = upper_bound(l.begin(), l.end(), r[i]) - l.begin();
            ans_r += suf_l[pos_r] - r[i] * (ll) (n-pos_r);
            mini = min(mini, ans_r);
        }
        FOR(i,0,n){
            ll ans_l = suf_l[i] - l[i] * (ll) (n-i);
            if (l[i] <= r[0]){
                mini = min(mini, ans_l);
                continue;
            }
            ll pos_l = upper_bound(r.begin(), r.end(), l[i]) - r.begin();
            ans_l += l[i] * (ll) (pos_l) - pre_r[pos_l-1];
            mini = min(mini, ans_l);
 
        }
        ans+=mini*2;
        cout<<ans;
        return;
    }
}
 
int main() {
    ios::sync_with_stdio(false);
    int TC = 1;
    //cin >> TC;
    FOR(i, 1, TC+1){
        //cout << "Case #" << i << ": ";
        solve();
    }
    return 0;
}

컴파일 시 표준 에러 (stderr) 메시지

bridge.cpp:10: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
   10 | #pragma GCC optimization ("Ofast")
      | 
bridge.cpp:11: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
   11 | #pragma GCC optimization ("unroll-loops")
      |

• Subtask 1: 0 / 8
• Subtask 2: 0 / 14
• Subtask 3: 0 / 9
• Subtask 4: 0 / 32
• Subtask 5: 0 / 37