제출 #1163732

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1163732		pan	Rainy Markets (CCO22_day1problem2)	C++20	25 / 25	414 ms	76648 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll b[1000005], p[1000005], u[1000005], le[1000005], ri[1000005], here[1000005];

int main()
{
	cin.tie(nullptr);
	cout.tie(nullptr);
	ios::sync_with_stdio(false);
	ll n;
	cin >> n;
	for (ll i=1; i<=n; ++i) cin >> b[i];
	for (ll i=1; i<n; ++i) cin >> p[i];
	for (ll i=1; i<n; ++i) cin >> u[i];
	// greedy to make max flow
	for (ll i=1; i<=n-1; ++i)
	{
		ll take1 = min(b[i], p[i]);
		b[i] -= take1; le[i] += take1; p[i] -= take1;
		ll take2 = min(u[i], p[i]);
		u[i] -= take2; here[i] += take2; p[i] -= take2;
		ll take3 = min(b[i + 1], p[i]);
		b[i + 1] -= take3; ri[i] += take3; p[i] -= take3;
		if (p[i] > 0) {cout << "NO\n"; return 0;}
	}
	ll ans = 0;
	// greedy to minimise cost
	for (ll i=n-1; i>=1; --i)
	{
		// move here to ri first
		ll take1 = min(here[i], b[i + 1]);
		b[i + 1] -= take1; here[i] -= take1; ri[i] += take1; 
		// then move le to ri
		ll take2 = min(le[i], b[i + 1]);
		b[i] += take2; b[i+1] -= take2, le[i] -= take2, ri[i] += take2;
		ans += here[i];
	}
	cout << "YES\n" << ans << endl;
	for (ll i=1; i<=n-1; ++i) cout << le[i] << ' ' << here[i] << ' ' << ri[i] << endl;
	
	
	return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 5 / 5
• Subtask 3: 6 / 6
• Subtask 4: 9 / 9