oj.uz

답안 #320073

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
320073 2020-11-07T13:08:35 Z mohamedsobhi777 Pinball (JOI14_pinball) C++14
0 / 100
 8 ms 13036 KB 
#include <bits/stdc++.h>

#pragma GCC optimize("-Ofast")
//#pragma GCC optimize("trapv")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-funroll-loops")

#define I inline void
#define S struct
#define vi vector<int>
#define vii vector<pair<int, int>>
#define pii pair<int, int>
#define pll pair<ll, ll>

using namespace std;
using ll = long long;
using ld = long double;

const int N = 2e5 + 7, mod = 1e9 + 7;
const ll inf = 2e18;

// How interesting!

int n, m;
map<int, int> mp;
ll cl[N], cr[N];
vector<int> ord;

struct dev
{
        ll l, md, r, c;
        dev() {}
        dev(int _l, int _md, int _r, int _c)
        {
                l = _l;
                md = _md;
                r = _r;
                c = _c;
        }
};
vector<dev> v;

struct segtree
{
        ll tree[4 * N];
        segtree()
        {
                fill(tree, tree + 4 * N, inf);
        }
        void update(int node, int L, int R, int ix, ll val)
        {
                if (L == R)
                {
                        tree[node] = min(tree[node], val);
                        return;
                }
                int mid = (L + R) >> 1;
                if (ix <= mid)
                        update(node * 2 + 1, L, mid, ix, val);
                else
                        update(node * 2 + 2, mid + 1, R, ix, val);
                tree[node] = min(tree[node * 2 + 1], tree[node * 2 + 2]);
        }

        ll query(int node, int L, int R, int l, int r)
        {
                if (l > R || r < L)
                        return inf;
                if (L >= l && R <= r)
                        return tree[node];
                int mid = (L + R) >> 1;
                ll s1 = query(node * 2 + 1, L, mid, l, r);
                ll s2 = query(node * 2 + 2, mid + 1, R, l, r);
                return min(s1, s2);
        }
        inline ll query(int l, int r) { return query(0, 1, N, l, r); }
        inline void upd(int pos, ll val) { update(0, 1, N, pos, val); }
} s[2];

int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);

        //freopen("in.in", "r", stdin);
        cin >> n >> m;
        for (int i = 1; i <= n; ++i)
        {
                int a, b, c, d;
                cin >> a >> b >> c >> d;
                v.push_back(dev(a, c, b, d));
                ord.push_back(a);
                ord.push_back(b);
                ord.push_back(c);
        }
        sort(ord.begin(), ord.end());
        ord.erase(unique(ord.begin(), ord.end()), ord.end());

        for (int i = 0; i < (int)ord.size(); ++i)
        {
                mp[ord[i]] = i + 1;
        }
        for (int i = 0; i < n; ++i)
        {
                v[i].l = mp[v[i].l];
                v[i].r = mp[v[i].r];
                v[i].md = mp[v[i].md];
        }

        for (int i = 0; i < n; ++i)
        {
                cl[i] = (v[i].l == 1 ? 0 : s[0].query(v[i].l, v[i].r)) + v[i].c;
                cr[i] = (v[i].r == mp[m] ? 0 : s[1].query(v[i].l, v[i].r)) + v[i].c;
                s[0].upd(v[i].md, cl[i]);
                s[1].upd(v[i].md, cr[i]);
        }

        ll ans = inf;
        for (int i = 0; i < n; ++i)
        {
                ans = min(ans, cl[i] + cr[i] - v[i].c);
        }
        if (ans == inf)
                cout << -1;
        else
                cout << ans;
        return 0;
}

/*
        - bounds sir (segtree = 4N, eulerTour = 2N, ...)
        - a variable defined twice?
        - will overflow?
        - is it a good complexity?
        - don't mess up indices (0-indexed vs 1-indexed)
        - reset everything between testcases. 
*/

Subtask #1
0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Correct 7 ms 12908 KB Output is correct
2 Correct 7 ms 12908 KB Output is correct
3 Correct 7 ms 12908 KB Output is correct
4 Correct 7 ms 12908 KB Output is correct
5 Correct 7 ms 13036 KB Output is correct
6 Correct 8 ms 12908 KB Output is correct
7 Incorrect 7 ms 12908 KB Output isn't correct
8 Halted 0 ms 0 KB -

Subtask #2
0 / 18.0

# 결과 실행 시간 메모리 Grader output
1 Correct 7 ms 12908 KB Output is correct
2 Correct 7 ms 12908 KB Output is correct
3 Correct 7 ms 12908 KB Output is correct
4 Correct 7 ms 12908 KB Output is correct
5 Correct 7 ms 13036 KB Output is correct
6 Correct 8 ms 12908 KB Output is correct
7 Incorrect 7 ms 12908 KB Output isn't correct
8 Halted 0 ms 0 KB -

Subtask #3
0 / 22.0

# 결과 실행 시간 메모리 Grader output
1 Correct 7 ms 12908 KB Output is correct
2 Correct 7 ms 12908 KB Output is correct
3 Correct 7 ms 12908 KB Output is correct
4 Correct 7 ms 12908 KB Output is correct
5 Correct 7 ms 13036 KB Output is correct
6 Correct 8 ms 12908 KB Output is correct
7 Incorrect 7 ms 12908 KB Output isn't correct
8 Halted 0 ms 0 KB -

Subtask #4
0 / 49.0

# 결과 실행 시간 메모리 Grader output
1 Correct 7 ms 12908 KB Output is correct
2 Correct 7 ms 12908 KB Output is correct
3 Correct 7 ms 12908 KB Output is correct
4 Correct 7 ms 12908 KB Output is correct
5 Correct 7 ms 13036 KB Output is correct
6 Correct 8 ms 12908 KB Output is correct
7 Incorrect 7 ms 12908 KB Output isn't correct
8 Halted 0 ms 0 KB -