P1527 [国家集训队]矩阵乘法

知识点: 整体二分

原题面 Luogu


题意简述

给定一 \(n \times n\) 的矩阵 \(a\)\(q\) 次询问。
每次询问一个给定子矩形的第 \(k\) 小数。
\(1\le n\le 500\)\(1\le q\le 6\times 10^4\)\(0\le a_{i,j}\le 10^9\)


分析题意

静态子矩阵第 k 小值。
整体二分经典应用,将区间第 k 小值中的树状数组改为二维树状数组即可。
区间第 k 小值:Link


爆零小技巧

不可这样写,因为 posy_ 会被赋值,导致循环无法进行。

int Sum(int posx_, int posy_) {
  int ret = 0;
  for (; posx_; posx_ -= lowbit(posx_)) {
    for (; posy_; posy_ -= lowbit(posy_)) {
      if (ti[posx_][posy_] < time) {
        t[posx_][posy_] = 0;
        ti[posx_][posy_] = time;
      }
      ret += t[posx_][posy_]; 
    }
  }
  return ret;
}

代码实现

//知识点:整体二分 
/*
By:Luckyblock
*/
#include <algorithm>
#include <cstdio>
#include <ctype.h>
#include <cstring>
#define ll long long
const int kMaxn = 500 + 10; 
const int kMaxq = 5e5 + 10;
//=============================================================
struct Operation {
  bool type;
  int x1, y1, x2, y2, k, id;
  int val, posx, posy;
} q[kMaxq], lq[kMaxq], rq[kMaxq];
int n, m, maxa, a[kMaxn][kMaxn], ans[kMaxn];
int data_num, data[kMaxn * kMaxn], map[kMaxn * kMaxn];
//=============================================================
inline int read() {
  int f = 1, w = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) w = (w << 3) + (w << 1) + (ch ^ '0');
  return f * w;
}
void GetMax(int &fir_, int sec_) {
  if (sec_ > fir_) fir_ = sec_;
}
void GetMin(int &fir_, int sec_) {
  if (sec_ < fir_) fir_ = sec_;
}
namespace TreeArray {
  #define lowbit(x) (x&-x)
  const int kMaxm = 500;
  int time, ti[kMaxm + 10][kMaxm + 10], t[kMaxm + 10][kMaxm + 10];
  void Clear() {
    time ++;
  }
  void Add(int posx_, int posy_, int val_) {
    for (int x = posx_; x <= n; x += lowbit(x)) {
      for (int y = posy_; y <= n; y += lowbit(y)) {
        if (ti[x][y] < time) {
          t[x][y] = 0;
          ti[x][y] = time;
        }
        t[x][y] += val_; 
      }
    }
  }
  int Sum(int posx_, int posy_) {
    int ret = 0;
    for (int x = posx_; x; x -= lowbit(x)) {
      for (int y = posy_; y; y -= lowbit(y)) {
        if (ti[x][y] < time) {
          t[x][y] = 0;
          ti[x][y] = time;
        }
        ret += t[x][y]; 
      }
    }
    return ret;
  }
} 
void Prepare() {
  n = read(), m = read();
  for (int i = 1; i <= n; ++ i) {
    for (int j = 1; j <= n; ++ j) {
      data[++ data_num] = a[i][j] = read();  
    }
  }
  std :: sort(data + 1, data + data_num + 1);
  int lth = std :: unique(data + 1, data + data_num + 1) - data - 1;
  for(int i = 1; i <= n; i ++) {
    for (int j = 1; j <= n; ++ j) {
      int ori = a[i][j];
      a[i][j] = std :: lower_bound(data + 1, data + lth + 1, ori) - data;
      map[a[i][j]] = ori;
      GetMax(maxa, a[i][j]);
    }
  }
}
#define mid ((l_+r_)>>1)
void Solve(int l_, int r_, int ql_, int qr_) {
  if (ql_ > qr_) return ;
  if (l_ == r_) {
    for (int i = ql_; i <= qr_; ++ i) {
      if (q[i].type) ans[q[i].id] = l_;
    }
    return ;
  }
  int pl = 0, pr = 0;
  for (int i = ql_; i <= qr_; ++ i) {
    if (! q[i].type) {
      if (q[i].val <= mid) {
        TreeArray :: Add(q[i].posx, q[i].posy, 1);
        lq[++ pl] = q[i];
      } else {
        rq[++ pr] = q[i];
      }
    } else {
      int ret = TreeArray :: Sum(q[i].x2, q[i].y2);
      ret -= TreeArray :: Sum(q[i].x2, q[i].y1 - 1);
      ret -= TreeArray :: Sum(q[i].x1 - 1, q[i].y2);
      ret += TreeArray :: Sum(q[i].x1 - 1, q[i].y1 - 1);
      if (q[i].k <= ret) {
        lq[++ pl] = q[i];
      } else {
        q[i].k -= ret;
        rq[++ pr] = q[i];
      }
    }
  }
  TreeArray :: Clear();
  for (int i = 1; i <= pl; ++ i) q[ql_ + i - 1] = lq[i];
  for (int i = 1; i <= pr; ++ i) q[ql_ + pl + i - 1] = rq[i];
  Solve(l_, mid, ql_, ql_ + pl - 1);
  Solve(mid + 1, r_, ql_ + pl, qr_);
}
//=============================================================
int main() {
  Prepare();
  for (int i = 1; i <= n; ++ i) {
    for (int j = 1; j <= n; ++ j) {
      q[n * (i - 1) + j] = (Operation) {false, 0, 0, 0, 0, 0, 0, a[i][j], i, j};  
    }
  }
  for (int i = 1; i <= m; ++ i) {
    int x1 = read(), y1 = read(), x2 = read(), y2 = read();
    q[n * n + i] = (Operation) {true, x1, y1, x2, y2, read(), i, 0, 0, 0}; 
  }
  Solve(1, maxa, 1, n * n + m);
  for (int i = 1; i <= m; ++ i) {
    printf("%d\n", map[ans[i]]);
  }
  return 0;
}
posted @ 2020-09-01 22:24  Luckyblock  阅读(201)  评论(0编辑  收藏  举报