Submission #1199089

#	Time	Username	Problem	Language	Result	Execution time	Memory
1199089		InvMOD	Fortune Telling 2 (JOI14_fortune_telling2)	C++17	100 / 100	259 ms	53216 KiB

fortune_telling2

#include<bits/stdc++.h>

using namespace std;

#define fi first
#define se second
#define pb push_back
#define eb emplace_back

#define vi vector<int>
#define pi pair<int,int>
#define sz(v) (int)(v).size()
#define all(v) (v).begin(), (v).end()
#define compact(v) (v).erase(unique(all(v)), (v).end())

template<class T> using upq = priority_queue<T, vector<T>, greater<T>>;
template<class T> int lwrbound(const vector<T>& a, const T& b, const int s = 0){return int(lower_bound(s + all(a), b) - a.begin());}
template<class T> int uprbound(const vector<T>& a, const T& b, const int s = 0){return int(upper_bound(s + all(a), b) - a.begin());}

#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 sumof(x) accumulate(all(x), 0ll)
#define dbg(x) "[" << #x " = " << (x) << "]"
#define el "\n"

using ll = long long;
using ld = long double;

template<class T> bool ckmx(T& a, const T b){return (a < b ? a = b, true : false);}
template<class T> bool ckmn(T& a, const T b){return (a > b ? a = b, true : false);}

const int N = 2e5 + 5;
const int MOD = 1e9 + 7;
const ll INF = numeric_limits<ll>::max() / 8;

/*
    a[i] <= b[i]

    Type 1: a[i] <= T[j] < b[i] -> swap only a[i] to b[i]
    Type 2: b[i] <= T[j] -> swap both a[i], b[i]

    a[i]..... Type 1.... b[i]
    b[i]..... Type 1.... b[i]
    -> we will find the last type 1 and calculate the answer
*/

struct SegTree{
    vector<int> st; int trsz;

    SegTree(int n = 0): st((n << 2) | 1), trsz(n) {}

    void update(int id, int l, int r, int x, int p){
        if(l == r){
            st[id] = max(st[id], p);
        }
        else{
            int m = l+r>>1;
            if(x <= m)
                update(id << 1, l, m, x, p);
            else update(id << 1|1, m+1, r, x, p);
            st[id] = max(st[id << 1], st[id << 1|1]);
        }
    }

    int get(int id, int l, int r, int u, int v){
        if(l > v || r < u) return 0;
        if(l >= u && r <= v) return st[id];

        int m = l+r>>1;
        return max(get(id << 1, l, m, u, v), get(id << 1|1, m+1, r, u, v));
    }

    void update(int x, int p){
        update(1, 0, trsz, x, p);
    }

    int query(int l, int r){
        if(l > r) return 0;
        return get(1, 0, trsz, l, r);
    }
};

struct IT{
    vector<vi> st; int trsz;

    IT(int n, vector<int>& a): st((n << 2) | 1), trsz(n){
        build(1, 0, n, a);
    }

    void merge_st(int id, int l, int r){
        int i = 0, j = 0;
        while(i < sz(st[l]) && j < sz(st[r])){
            if(st[l][i] < st[r][j]){
                st[id].eb(st[l][i]);
                i++;
            }
            else{
                st[id].eb(st[r][j]);
                j++;
            }
        }
        while(i < sz(st[l])) st[id].eb(st[l][i++]);
        while(j < sz(st[r])) st[id].eb(st[r][j++]);
    }

    void build(int id, int l, int r, vector<int>& a){
        if(l == r){
            st[id].eb(a[l]);
        }
        else{
            int m = l+r>>1;
            build(id << 1, l, m, a);
            build(id << 1|1, m+1, r, a);
            merge_st(id, id << 1, id << 1|1);
        }
    }

    int query(int l, int r, int k){
        return query(1, 0, trsz, l, r, k);
    }

    int query(int id, int l, int r, int u, int v, int k){
        if(l >= u && r <= v) return (sz(st[id]) - lwrbound(st[id], k));

        int m = l+r>>1;
        return (u <= m ? query(id << 1, l, m, u, v, k) : 0) +
               (v > m ? query(id << 1|1, m+1, r, u, v, k) : 0);
    }
};

void Main()
{
    int n,k; cin >> n >> k;

    vi a(n + 1), b(n + 1);
    FOR(i, 1, n) cin >> a[i] >> b[i];

    vi Q(k + 1); FOR(i, 1, k) cin >> Q[i];

    vi comp(1, -1);
    FOR(i, 1, k) comp.eb(Q[i]);

    sort(all(comp)), compact(comp);

    SegTree st(sz(comp)); IT tr(k, Q);
    FOR(i, 1, k){
        Q[i] = lwrbound(comp, Q[i]);
        st.update(Q[i], i);
    }

    ll ans = 0;
    FOR(i, 1, n){
        int x = lwrbound(comp, min(a[i], b[i]));
        int y = lwrbound(comp, max(a[i], b[i]));

        int lst = st.query(x, y - 1);
        int change = tr.query(lst + 1, k, max(a[i], b[i]));

        if(!lst){
            ans += ((change & 1) ? b[i] : a[i]);
        }
        else{
            ans += ((change & 1) ? min(a[i], b[i]) : max(a[i], b[i]));
        }
    }
    cout << ans << el;
}

int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    #define name "InvMOD"
    if(fopen(name".INP", "r")){
        freopen(name".INP", "r", stdin);
        freopen(name".OUT", "w", stdout);
    }

    int t = 1; while(t--) Main();
    return 0;
}

Compilation message (stderr)

fortune_telling2.cpp: In function 'int32_t main()':
fortune_telling2.cpp:176:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  176 |         freopen(name".INP", "r", stdin);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~
fortune_telling2.cpp:177:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  177 |         freopen(name".OUT", "w", stdout);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 4 / 4
• Subtask 2: 31 / 31
• Subtask 3: 65 / 65