結果
問題 | No.448 ゆきこーだーの雨と雪 (3) |
ユーザー |
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提出日時 | 2019-03-03 17:33:48 |
言語 | C++11 (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 171 ms / 2,000 ms |
コード長 | 2,062 bytes |
コンパイル時間 | 807 ms |
コンパイル使用メモリ | 79,008 KB |
実行使用メモリ | 9,676 KB |
最終ジャッジ日時 | 2024-06-23 13:16:18 |
合計ジャッジ時間 | 5,072 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 35 |
ソースコード
#include <iostream>#include <algorithm>#include <cstring>#include <sstream>#include <map>#include <queue>#include <set>#include <vector>#include <stack>#include <cstdio>#include <cmath>#define mk make_pair#define pb push_back#define printf printf_s#define scanf scanf_susing namespace std;typedef long long int ll;typedef pair<ll, ll> pos;const ll MOD = 1000000007, N = 200001, dx[4] = { -1,1,0,0 }, dy[4] = { 0,0,-1,1 }, MIN = -72340172838, MAX = 72340172838;ll mn(ll a, ll b) {if (a == -1)return b;if (b == -1)return a;return min(a, b);}ll gcd(ll v1, ll v2) {if (v1 == 0)return v2; if (v2 == 0)return v1; if (v2 > v1)return gcd(v2%v1, v1); return gcd(v1%v2, v2);}ll pw(ll v1, ll v2) {ll v3 = 1;while (v2 > 0) {if (v2 % 2)v3 = (v3*v1) % MOD;v1 = (v1*v1) % MOD;v2 /= 2;}return v3;}struct ab {ll a, b;bool operator<(const ab& right) const {return right.b*a - right.a*b < 0;}};ll n, k, mx, mnv, ans = 0;pos pr[N];queue<ll> ql;bool ok(ll md) {ll bt = MIN;for (int i = 0; i < n; i++) {if (pr[i].second <= md)continue;if (pr[i].first - bt < k)return false;bt = pr[i].first;}return true;}int main() {cin >> n >> k;for (int i = 0; i < n; i++) { cin >> pr[i].first >> pr[i].second; mx = max(mx, pr[i].second); ans += pr[i].second; }if (ok(0)) { cout << 0 << endl << 0 << endl; return 0; }while (mx - mnv > 1) {ll md = (mx + mnv) / 2;if (ok(md))mx = md;else mnv = md;}cout << mx << endl;for (int i = 0; i < n; i++) { if (pr[i].second > mx) { ql.push(pr[i].first); ans -= pr[i].second; } }ql.push(MAX);ll dp[N] = {}, mxv[N] = {},lt=MIN,bt=-1;for (int i = 0; i < n; i++) {while (ql.front() < pr[i].first) { lt = ql.front(); ql.pop(); }if (pr[i].first - lt >= k && ql.front() -pr[i].first >= k) {while (pr[i].first - pr[bt + 1].first>=k)bt++;ll btv = 0;if (bt != -1)btv = mxv[bt];dp[i] = btv + pr[i].second;}if (i != 0)mxv[i] = max(dp[i], mxv[i - 1]);else mxv[i] = dp[i];}cout << ans - mxv[n - 1] << endl;return 0;}