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;
}
作者@Luckyblock,转载请声明出处。