QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#417186#8718. 保区间最小值一次回归问题whxqwqTL 8ms59112kbC++142.7kb2024-05-22 15:53:212024-05-22 15:53:23

Judging History

你现在查看的是最新测评结果

  • [2024-05-22 15:53:23]
  • 评测
  • 测评结果:TL
  • 用时:8ms
  • 内存:59112kb
  • [2024-05-22 15:53:21]
  • 提交

answer

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
void read (int &x) {
    char ch = getchar(); x =0 ; while (!isdigit(ch)) ch =getchar();
    while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
} const int N = 5e5 + 5;
int n, m, a[N];
vector<int> del[N], add[N];
#define pb push_back
struct qwq { int l, r, v; } b[N];
bool cmp (qwq x, qwq y) {
    return x.v == y.v ? x.l < y.l : x.v < y.v;
}
multiset <int> s;
multiset <int>::iterator it;
int val[N];
vector<int> w[N];
vector<pair<int, int> > seg[N]; 
#define ll long long
int L[N];
multiset<ll> u;  ll f[N];
multiset<ll>::iterator iu;
ll calc (int x) {
    int lim = 0;
    // printf ("666 %d\n", val[x]);
    // for (auto x : seg[x]) printf ("%d %d\n", x.first, x.second);
    // for (int y : w[x]) printf ("%d ", y); puts ("");
    for (auto x : seg[x]) L[x.second] = max (L[x.second], x.first);
    for (auto x : seg[x]) lim = max (lim, x.first);
    int num = (int)w[x].size(), la = 0;
    ll ans = 1e18;
    u.insert (0);
    for (int i = 1; i <= num; ++i) {
        int now = w[x][i - 1];
        // printf ("%d %d\n", now, )
        f[i] = *u.begin() + abs (a[now] - val[x]);
        u.insert (f[i]);
        if (L[now] > 0) u.erase (0);
        while (la < num && w[x][la] < L[now]) {
            iu = u.find (f[la + 1]); u.erase (iu); ++la;
        }
        if (now >= lim) ans = min (ans, f[i]);
    }
    // printf ("hhh %lld\n", ans);
    u.clear();
    for (auto x : seg[x]) L[x.second] = 0;
    return ans;
}
void work () {
    read (n), read (m);
    for (int i = 1; i <= n; ++i) read (a[i]);
    for (int i = 1; i <= n; ++i) del[i].clear(), add[i].clear();
    for (int i = 1, l, r, v; i <= m; ++i) {
        read (l), read (r), read (v);
        b[i] = (qwq){l, r, v}; val[i] = v;
    }
    sort (val + 1, val + m + 1);
    int tot = unique (val + 1, val + m + 1) - val - 1;
    for (int i = 1; i <= tot; ++i) seg[i].clear(), w[i].clear();
    for (int i = 1; i <= m; ++i) {
        int v = lower_bound (val + 1, val + m + 1, b[i].v) - val;
        b[i].v = v;
        add[b[i].l].pb (v), del[b[i].r + 1].pb (v);
        seg[v].push_back (make_pair (b[i].l, b[i].r));
    }
    s.clear ();
    for (int i = 1; i <= n; ++i) {
        for (int x : add[i]) s.insert (-x);
        for (int x : del[i]) it = s.find (-x), s.erase (it);
        if ((int)s.size() == 0) continue;
        int now = *s.begin();
        w[-now].push_back (i);
    }   
    long long ans = 0;
    for (int i = 1; i <= tot; ++i) ans += calc (i);
    printf ("%lld\n", ans);
}
signed main() {
    int T; read (T);
    for (int i = 1; i <= T && i <= 5; ++i) work ();
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 8ms
memory: 59112kb

input:

1
3 2
2023 40 41
1 1 2022
2 3 39

output:

2

result:

ok 1 number(s): "2"

Test #2:

score: -100
Time Limit Exceeded

input:

1000
100 100
1 35141686 84105222 84105220 7273527 178494861 178494861 112519027 77833654 77833656 261586535 278472336 278472336 261586536 416361017 416361017 426649080 323519513 278472337 420127821 420127823 420127823 482516531 434108818 420127821 631535744 615930922 546346921 546346920 546346920 70...

output:


result: