結果

問題 No.1014 competitive fighting
コンテスト
ユーザー Sumitacchan
提出日時 2020-03-20 22:31:02
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
(最新)
AC  
(最初)
実行時間 -
コード長 5,131 bytes
コンパイル時間 1,998 ms
コンパイル使用メモリ 179,848 KB
実行使用メモリ 16,768 KB
最終ジャッジ日時 2024-07-07 12:52:28
合計ジャッジ時間 10,513 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 50 WA * 1
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <bits/stdc++.h>
using namespace std;
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<typename T> using gpp_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T, typename L> using gpp_map = tree<T, L, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T> using gpp_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;*/
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define bit_exist(x, n) ((x >> n) & 1)
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) { cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl; }
#define mdebug(m) { cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;} }
#define Return(ans) { cout << (ans) << endl; return 0; }
#define pb push_back
#define f first
#define s second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p){ cout << '(' << p.first << ',' << p.second << ')'; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void Ans(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;
//#define Add(x, y) x = (x + (y)) % mod
//#define Mult(x, y) x = (x * (y)) % mod



struct SegmentTree
{
    using T = int;

    int N;
    vector<T> dat;

    T id = INF;
    T F(T &a, T &b) { return min(a, b); }
    
    SegmentTree(int n){
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<T>(2 * N - 1, id);
    }

    SegmentTree(int n, vector<T> &v){
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<T>(2 * N - 1, id);
        for(int i = 0; i < n; i++) dat[i + N - 1] = v[i];
        for(int i = N - 2; i >= 0; i--) dat[i] = F(dat[i * 2 + 1], dat[i * 2 + 2]); 
    }

    SegmentTree(){}


    void update(int k, T a){
        k += N - 1;
        dat[k] = a;
        while(k > 0){
            k = (k - 1) / 2;
            dat[k] = F(dat[k * 2 + 1], dat[k * 2 + 2]);
        }
    }

    void reset() { fill(dat.begin(), dat.end(), id); }

    T get(int a, int b, int k, int l, int r){
        if(r <= a || b <= l) return id;
        if(a <= l && r <= b) return dat[k];
        else{
            T vl = get(a, b, k * 2 + 1, l, (l + r) / 2);
            T vr = get(a, b, k * 2 + 2, (l + r) / 2, r);
            return F(vl, vr);
        }
    }
    T get(int a, int b) { return get(a, b, 0, 0, N); }

    T val(int k){ return dat[k + N - 1]; }
};

signed main(){

    int N; cin >> N;
    vec A(N), B(N), C(N);
    pairs p_bc(N);
    REP(i, N){
        cin >> A[i] >> B[i] >> C[i];
        p_bc[i] = Pii(B[i] - C[i], i);
    }
    Sort(p_bc);
    vec inv_bc(N);
    REP(i, N) inv_bc[p_bc[i].s] = i;

    SegmentTree st(N);
    int m = INF;
    REP(i, N){
        int i0 = Lower_bound(p_bc, Pii(A[i], -INF));
        int ma = st.get(i0, N);
        if(ma <= B[i] - C[i]){
            //debug(i); debug(ma);
            Min(m, A[i]);
        }
        st.update(inv_bc[i], A[i]);
    }

    //debug(m);
    pairs p_abc(N), p_a(N);
    REP(i, N){
        p_abc[i] = Pii(B[i] - C[i] + A[i], i);
        p_a[i] = Pii(A[i], i);
    }
    Sort(p_abc);
    Sort(p_a);
    vec inv_a(N);
    REP(i, N) inv_a[p_a[i].s] = i;

    SegmentTree ST(N);
    vec ans(N);
    REP(_, N){
        int i = p_abc[_].s;
        //debug(i);

        if(B[i] - C[i] >= m){
            ans[i] = -1;
            continue;
        }

        int i0 = Upper_bound(p_a, Pii(B[i] - C[i], INF));
        ans[i] = max(0LL, -ST.get(0, i0)) + B[i];

        ST.update(inv_a[i], -ans[i]);
    } 
    REP(i, N) Ans(ans[i] == -1, "BAN", ans[i]);

    return 0;
}
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