CodeForces 1221F Choose a Square

pufanyi

发布于：Oct 4, 2019

题意大概就是有 $n$ 个点，每个点其坐标 $x_i,y_i$ 与权值 $c_i$ ，其中 $1\le n\le 5\cdot 10^5,0\le x_i,y_i\le 10^9,-10^6\le c_i\le 10^6$ 。

让你选一个正方形，该正方形的左下角及右上角必须在 $y=x$ 这条直线上。所获得的权值为在正方形内的点的权值和减去正方形的边权。输出所获的最大权值及其选择正方形的左下角 $x_1,y_1$ 及右上角 $x_2,y_2$ ，要求 $0\le x_1=y_1\le x_2=y_2\le 2\cdot 10^9$ 。

其实就是让你选两个数 $l,r$ ，左下角为 $(l,l)$ ，右上角为 $(r,r)$ 。

我们考虑一个点是否在正方形内。

我们发现对于第 $i$ 个点，若 $l\le \min\{x_i,y_i\}\le \max\{x_i,y_i\}\le r$ ，那么 $(x_i,y_i)$ 就在正方形内。

所以我们发现答案就是：

\sum_{\max\{x_i,y_i\}\le r} c_i-\sum_{\min\{x_i,y_i\}< l\le \max\{x_i,y_i\}\le r}c_i-(r-l+1)

我们考虑枚举右端点 $r$ ，也就是枚举 $\sum_{\max\{x_i,y_i\}\le r} c_i-r$ ，同时用线段树维护 $-\sum_{\min\{x_i,y_i\}< l\le \max\{x_i,y_i\}\le r}c_i+l-1$ 的最小值即可。

#define _CRT_SECURE_NO_WARNINGS

#include <map>
#include <set>
#include <stack>
#include <ctime>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cctype>
#include <vector>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <fstream>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

inline char gc() {
  static const LL L = 233333;
  static char sxd[L], *sss = sxd, *ttt = sxd;
  if (sss == ttt) {
    ttt = (sss = sxd) + fread(sxd, 1, L, stdin);
    if (sss == ttt) {
      return EOF;
    }
  }
  return *sss++;
}

#ifndef dd
#define dd c = gc()
#endif
inline char readalpha() {
  char dd;
  for (; !isalpha(c); dd);
  return c;
}

inline char readchar() {
  char dd;
  for (; c == ' '; dd);
  return c;
}

template <class T>
inline bool read(T& x) {
  bool flg = false;
  char dd;
  x = 0;
  for (; !isdigit(c); dd) {
    if (c == '-') {
      flg = true;
    } else if(c == EOF) {
      return false;
    }
  }
  for (; isdigit(c); dd) {
    x = (x << 1) + (x << 3) + (c ^ 48);
  }
  if (flg) {
    x = -x;
  }
  return true;
}
#undef dd

template <class T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x < 10) {
    putchar(x | 48);
    return;
  }
  write(x / 10);
  putchar((x % 10) | 48);
}

typedef long long LL;

const LL maxn = 1000005;

LL n;
LL _cnt = 0;

set<LL> mj;

#define ls(x) (x << 1)
#define rs(x) (x << 1 | 1)

struct Tree {
  struct Node {
    pair<LL, LL> xx;
    LL lzy;
  } no[maxn << 2];

  inline void push_up(LL x) {
    no[x].xx = max(no[ls(x)].xx, no[rs(x)].xx);
  }

  inline void build_tree(LL l, LL r, LL k) {
    static auto x = mj.begin();
    if (l == r) {
      auto tmp = x;
      ++tmp;
      if (tmp != mj.end()) {
        no[k].xx = make_pair(*tmp, *x);
      } else {
        no[k].xx = make_pair(-233333, *x);
      }
      x++;
      return;
    }
    LL mid = (l + r) >> 1;
    build_tree(l, mid, ls(k));
    build_tree(mid + 1, r, rs(k));
    push_up(k);
  }

  inline void push_down(LL k) {
    if (no[k].lzy) {
      no[ls(k)].xx.first -= no[k].lzy;
      no[rs(k)].xx.first -= no[k].lzy;
      no[ls(k)].lzy += no[k].lzy;
      no[rs(k)].lzy += no[k].lzy;
      no[k].lzy = 0;
    }
  }

  inline void add(LL l, LL r, LL k, LL L, LL R, LL x) {
    if (L <= l && r <= R) {
      no[k].lzy += x;
      no[k].xx.first -= x;
      return;
    }
    LL mid = (l + r) >> 1;
    push_down(k);
    if (L <= mid) {
      add(l, mid, ls(k), L, R, x);
    }
    if (R > mid) {
      add(mid + 1, r, rs(k), L, R, x);
    }
    push_up(k);
  }

  inline pair<LL, LL> query(LL l, LL r, LL k, LL L, LL R) {
    if (L <= l && r <= R) {
      return no[k].xx;
    }
    LL mid = (l + r) >> 1;
    push_down(k);
    if (R <= mid) {
      return query(l, mid, ls(k), L, R);
    } else if (L > mid) {
      return query(mid + 1, r, rs(k), L, R);
    } else {
      return max(query(l, mid, ls(k), L, R), query(mid + 1, r, rs(k), L, R));
    }
  }
} tr;

struct QJ {
  LL mn, mx, mnid, mxid, qz;

  friend bool operator < (QJ a, QJ b) {
    return a.mx < b.mx;
  }
} qj[maxn];

struct LS {
  LL x, id;

  friend bool operator < (LS a, LS b) {
    return a.x < b.x;
  }
} ls[maxn << 1];

map<int, int> anss;

int main() {
  read(n);
  for (LL i = 1; i <= n; ++i) {
    LL x, y;
    read(x), read(y), read(qj[i].qz);
    if (x == y) {
      anss[x] += qj[i].qz;
    }
    qj[i].mn = min(x, y);
    qj[i].mx = max(x, y);
    ls[++_cnt].x = qj[i].mn;
    ls[_cnt].id = i << 1;
    ls[++_cnt].x = qj[i].mx;
    ls[_cnt].id = i << 1 | 1;
    mj.insert(x);
    mj.insert(y);
  }
  sort(ls + 1, ls + _cnt + 1);
  LL cnt = 0;
  for (LL i = 1; i <= _cnt; ++i) {
    if (i == 1 || ls[i].x != ls[i - 1].x) {
      cnt++;
    }
    if (ls[i].id & 1) {
      qj[ls[i].id >> 1].mxid = cnt;
    } else {
      qj[ls[i].id >> 1].mnid = cnt;
    }
  }
  sort(qj + 1, qj + n + 1);
  LL ansx = 1300000000, ansy = 1300000000, ans = 0;
  for (auto x : anss) {
    if (x.second > ans) {
      ans = x.second;
      ansx = ansy = x.first;
    }
  }
  tr.build_tree(1, cnt, 1);
  LL now = 1;
  LL __cnt = 0;
  LL sum = 0;
  int bg = *mj.begin();
  for (auto x : mj) {
    __cnt++;
    while (now <= n && qj[now].mx <= x) {
      tr.add(1, cnt, 1, qj[now].mnid, cnt, qj[now].qz);
      sum += qj[now].qz;
      now++;
    }
    if (__cnt > 1) {
      pair<LL, LL> an = tr.query(1, cnt, 1, 1, __cnt - 1);
      LL Ans = sum + an.first - x;
      if (Ans > ans) {
        auto tx = mj.lower_bound(an.second);
        ansx = *(++tx);
        ansy = x;
        ans = Ans;
      }
    }
    int len = x - bg;
    if (sum - len > ans) {
      ansx = bg, ansy = x, ans = sum - len;
    }
  }
  printf("%lld\n", ans);
  printf("%lld %lld %lld %lld\n", ansx, ansx, ansy, ansy);
  return 0;
}

更新于：Dec 26, 2023

CodeForces

数据结构

这个毒瘤的 live2d

来一波搞事。大概是这里：https://github.com/pufanyi/live

造一棵树

帮人造了几组数据。其实就是造了一棵树。记录一下以备后用。 12345678910111213141516171819202122232425262728293031323334353...