제출 #1040194

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1040194		BackNoob	Salesman (IOI09_salesman)	C++14	80 / 100	622 ms	37664 KiB

salesman

#include <bits/stdc++.h>
#define ll long long
#define fi first
#define se second
#define endl '\n'
#define MASK(i) (1LL << (i))
#define ull unsigned long long
#define ld long double
#define pb push_back
#define all(x) (x).begin() , (x).end()
#define BIT(x , i) ((x >> (i)) & 1)
#define TASK "task"
#define sz(s) (int) (s).size()
 
using namespace std;
const int mxN = 5e5 + 227;
const int inf = 1e9 + 277;
const int mod = 1e9 + 7;
const ll infll = 1e18 + 7;
const int base = 307;
const int LOG = 20;
 
template <typename T1, typename T2> bool minimize(T1 &a, T2 b) {
    if (a > b) {a = b; return true;} return false;
}
template <typename T1, typename T2> bool maximize(T1 &a, T2 b) {
    if (a < b) {a = b; return true;} return false;
}

struct Segtree{
    int n;
    vector<ll> t;
    Segtree(){}
    Segtree(int _n) {
        n = _n;
        int offset = 1;
        while(offset <= n) offset *= 2;
        t.resize(offset * 2 + 7, -infll);
    }

    void update(int v, int tl, int tr, int pos, ll val) {
        if(tl == tr) t[v] = val;
        else {
            int tm = (tl + tr) / 2;
            if(pos <= tm) update(v * 2, tl, tm, pos, val);
            else update(v * 2 + 1, tm + 1, tr, pos, val);
            t[v] = max(t[v * 2], t[v * 2 + 1]);
        }
    }

    ll getmax(int v, int tl, int tr, int l, int r) {
        if(l > r) return -infll;
        if(tl == l && tr == r) return t[v];
        int tm = (tl + tr) / 2;
        ll m1 = getmax(v * 2, tl, tm, l, min(r, tm));
        ll m2 = getmax(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
        return max(m1, m2);
    }

    void update(int pos, ll val) {
        update(1, 1, n, pos, val);
    }
    ll getmax(int l, int r) {
        return getmax(1, 1, n, l, r);
    }
} segl, segr;

struct item{
    int t, p, c;

    bool operator<(const item &other) const {
        if(t == other.t) return p < other.p;
        return t < other.t;
    }
} a[mxN];

int n;
int down;
int up;
int home;
ll dp[mxN];
 
void solve()
{
    cin >> n >> down >> up >> home;
    for(int i = 1; i <= n; i++) cin >> a[i].t >> a[i].p >> a[i].c;

    sort(a + 1, a + 1 + n);

    vector<int> val;
    val.pb(home);
    for(int i = 1; i <= n; i++) val.pb(a[i].p);
    sort(all(val));
    val.erase(unique(all(val)), val.end());

    segl = Segtree(sz(val));
    segr = Segtree(sz(val));
    memset(dp, -0x3f, sizeof(dp));
    int pos = lower_bound(all(val), home) - val.begin() + 1;
    segl.update(pos, home * up);
    segr.update(pos, -home * down);


    for(int i = 1; i <= n; i++) {
        int j = i;
        while(j <= n && a[i].t == a[j].t) ++j;

        ll curmax = dp[0];
        ll curmax2 = dp[0];
        ll pre = dp[0];
        for(int k = i; k < j; k++) {
            int pos = lower_bound(all(val), a[k].p) - val.begin() + 1;
            maximize(curmax, segl.getmax(1, pos - 1) - 1LL * a[k].p * up);
            if(k != i) {
                int lef = lower_bound(all(val), a[k - 1].p) - val.begin() + 1;
                maximize(curmax2, segr.getmax(lef, pos) + pre - 1LL * a[k].p * up);
            }
            curmax += a[k].c;
            curmax2 += a[k].c;
            pre += a[k].c;
            maximize(dp[k], curmax);
            maximize(dp[k], curmax2);
            maximize(pre, 1LL * a[k].p * down + 1LL * a[k].p * up + a[k].c);
            if(k + 1 < n) {
                curmax -= (a[k + 1].p - a[k].p) * up;
                curmax2 -= (a[k + 1].p - a[k].p) * up;
            }            
        }

        curmax = dp[0];
        curmax2 = dp[0];
        pre = dp[0];
        for(int k = j - 1; k >= i; k--) {
            int pos = lower_bound(all(val), a[k].p) - val.begin() + 1;
            maximize(curmax, segr.getmax(pos + 1, sz(val)) + 1LL * a[k].p * down);
            if(k != j - 1) {
                int rig = lower_bound(all(val), a[k + 1].p) - val.begin() + 1;
                maximize(curmax2, segl.getmax(pos, rig) + pre + 1LL * a[k].p * down);
            }
            curmax += a[k].c;
            curmax2 += a[k].c;
            pre += a[k].c;
            maximize(dp[k], curmax);
            maximize(dp[k], curmax2);
            maximize(pre, - 1LL * a[k].p * down - 1LL * a[k].p * up + a[k].c);
            if(k - 1 >= i) {
                curmax -= (a[k].p - a[k - 1].p) * down;
                curmax2 -= (a[k].p - a[k - 1].p) * down;
            }
        }
        for(int k = i; k < j; k++) {
            int pos = lower_bound(all(val), a[k].p) - val.begin() + 1;
            segl.update(pos, dp[k] + 1LL * a[k].p * up);
            segr.update(pos, dp[k] - 1LL * a[k].p * down);
        }
        i = j - 1;
    }

    ll res = 0;
    for(int i = 1; i <= n; i++) {
        if(a[i].p <= home) maximize(res, dp[i] - 1LL * (home - a[i].p) * up);
        else maximize(res, dp[i] - 1LL * (a[i].p - home) * down);
    }
    cout << res << endl;
}
 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    // freopen(TASK".out" , "w" , stdout);
 
    int tc = 1;
    //cin >> tc;
    while(tc--) {
        solve();
    }
    return 0;
}

• Subtask 1: 80 / 100