#28765 (C++11) No.59 鉄道の旅

• 問題: No.59 鉄道の旅
• ユーザー: koba-e964
• 提出日時: 2015-05-25 21:53:18
• 言語: C++11 (gcc 13.3.0)
• 結果: AC
• 実行時間: 73 ms / 5,000 ms
• コード長: 2,254 bytes
• コンパイル時間: 763 ms
• コンパイル使用メモリ: 91,240 KB
• 実行使用メモリ: 11,392 KB
• 最終ジャッジ日時: 2024-12-24 20:38:10
• 合計ジャッジ時間: 1,793 ms
• ジャッジサーバーID （参考情報）: judge4 / judge2

ファイルパターン	結果
sample	AC * 4
other	AC * 12

ソースコード

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#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#define REP(i,s,n) for(int i=(int)(s);i<(int)(n);i++)
using namespace std;
typedef long long int ll;
typedef vector<int> VI;
typedef pair<int, int> PI;
const double EPS=1e-9;
/**
 * Segment Tree. This data structure is useful for fast folding on intervals of an array
 * whose elements are elements of monoid M. Note that constructing this tree requires the identity
 * element of M and the operation of M.
 */
template<class I, class BiOp = I (*) (I, I)>
class SegTree {
    int n;
    vector<I> dat;
    BiOp op;
    I e;
public:
    SegTree(int n_, BiOp op, I e) : op(op), e(e) {
        n = 1;
        while (n < n_) { n *= 2; } // n is a power of 2
        dat.resize(2 * n);
        for (int i = 0; i < 2 * n - 1; i++) {
            dat[i] = e;
        }
    }
    /* ary[k] <- v */
    void update(int k, I v) {
        k += n - 1;
        dat[k] = v;
        while (k > 0) {
            k = (k - 1) / 2;
            dat[k] = op(dat[2 * k + 1], dat[2 * k + 2]);
        }
    }
    void update_array(int k, int len, I *vals) {
        for (int i = 0; i < len; ++i) {
            update(k + i, vals[i]);
        }
    }
    /* l,r are for simplicity */
    I querySub(int a, int b, int k, int l, int r) {
        // [a,b) and  [l,r) intersects?
        if (r <= a || b <= l) return e;
        if (a <= l && r <= b) return dat[k];
        I vl = querySub(a, b, 2 * k + 1, l, (l + r) / 2);
        I vr = querySub(a, b, 2 * k + 2, (l + r) / 2, r);
        return op(vl, vr);
    }
    /* [a, b] (note: inclusive) */
    I query(int a, int b) {
        return querySub(a, b + 1, 0, 0, n);
    }
};
int main(void){
  const int M = 1048500;
  int n, k;
  cin >> n >> k;
  SegTree<int, plus<int> > st(M, plus<int>(), 0);
  REP (loop_var, 0, n) {
    int w; cin >> w;
    if (w > 0) {
      int q = st.query(w, M);
      if (q < k) {
    st.update(w, st.query(w, w) + 1);
      }
    } else {
      w = -w;
      int v = st.query(w, w);
      if (v > 0) {
    st.update(w, v - 1);
      }
    }
  }
  cout << st.query(1, M) << endl;
}
 
 
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