제출 #1203327

#제출 시각아이디문제언어결과실행 시간메모리
1203327MkswllPotatoes and fertilizers (LMIO19_bulves)C++20
100 / 100
102 ms12676 KiB
// I will become better someday.

#include <bits/stdc++.h>
using namespace std;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef pair <ll, int> pli;
typedef pair <int, ll> pil;
typedef pair <ll, ll> pll;
typedef pair <ld, ld> pdd;
#define debug(x) cout << '[' << #x << ": " << x << "] "
#define cio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define cases int _; cin >> _; while(_--)
#define pb push_back
#define eb emplace_back
#define space << " " <<
#define lb lower_bound
#define ub upper_bound
#define F first
#define S second
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define Unique(v) v.erase(unique(all(v)), v.end())
#define mset(x) memset(x, 0, sizeof(x))
#define sflush fflush(stdout)
#define cflush cout.flush()
#define yes cout << "YES\n"
#define no cout << "NO\n"
#define lyes cout << "Yes\n"
#define lno cout << "No\n"
#define nl cout << "\n";
#define vt vector
#define ar array
#define uid uniform_int_distribution 

template <typename T> 
istream& operator >> (istream& in, vector<T>& a){
	for(auto &x : a) in >> x; 
	return in;
}

template <typename T> 
ostream& operator << (ostream& out, vector<T>& a){
	for(auto &x : a) out << x << ' '; 
	return out;
}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());

ll read(){
    int w = 1, c;
    ll ret;
    while((c = getchar()) > '9' || c < '0'){
        w = (c == '-' ? -1 : 1);
    }
    ret = c - '0';
    while((c = getchar()) >= '0' && c <= '9'){
        ret = ret * 10 + c - '0';
    }
    return ret * w;
}

ll rd(){
	ll in;
	cin >> in;
	return in;
}
 
void write(ll x){
    if(x < 0){
        putchar('-');
        x = -x;
    }
    if(x > 9){
        write(x / 10);
    }
    putchar(x % 10 + '0');
}

const int MAXN = 5e5 + 5, MAXM = 2e5 + 5, INF = 1e9 + 5, MOD = 1e9 + 7;
const ll LMOD = (ll) 1e18 + 9;
const ll LINF = 1e18 + 5;
const ld ep = 1e-8, Pi = acos(-1.0);

int n, m, k, x; 
int a[MAXN], b[MAXN];
string s;

ll d[MAXN];

void clear(){
    
}

int main(){
    cio;
    cin >> n;
    priority_queue <ll> pq;
    ll ans = 0;
    for(int i = 1; i <= n; ++i){
    	cin >> a[i] >> b[i];
    	d[i] = d[i - 1] + a[i] - b[i];
    	
    }
    for(int i = 1; i <= n; ++i){
    	if(d[i] > d[n]){
    		ans += d[i] - d[n];
    		d[i] = d[n];
    	}
    	else if(d[i] < 0){
    		ans -= d[i];
    		d[i] = 0;
    	}
    	pq.push(d[i]);
    	pq.push(d[i]);
    	ans += pq.top() - d[i];
    	pq.pop();
    }
    
    cout << ans << "\n";
    return 0;
}   
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