BZOJ 2738 “矩阵乘法”（子矩阵第k大）

题意

给出一个矩阵，多次询问子矩阵中第k大的数是多少。

题解

我做这道题之前先照着这道题出了一道题，是这道题的一维版本，在这里：https://vijos.org/d/contest/p/5a26541bd3d8a11cef1706aa。

思路是这样的：二分答案mid，将所有小于mid的位置都在树状数组上 +1，对于每个询问，如果子矩阵所有”+1“之和 >= 这个询问的k，则把询问划分到”左边那一组”，否则划分到“右边那一组”，之后对它们分别处理。

核心代码：

void solve(int l, int r, int ql, int qr){
if(ql > qr) return;
if(l == r){
for(int i = ql; i <= qr; i++)
ans[q[i].id] = a[l].val;
return;
}
int mid = (l + r) >> 1, cnt = 0;
for(int i = l; i <= mid; i++)
add(a[i].x, a[i].y, 1);
for(int i = ql; i <= qr; i++){
sum[i] = ask(q[i]);
if(sum[i] >= q[i].k) cnt++;
}
for(int i = ql, pl = ql, pr = ql + cnt; i <= qr; i++)
if(sum[i] >= q[i].k) tmp[pl++] = q[i];
else q[i].k -= sum[i], tmp[pr++] = q[i];
for(int i = ql; i <= qr; i++) q[i] = tmp[i];
for(int i = l; i <= mid; i++)
add(a[i].x, a[i].y, -1);
solve(l, mid, ql, ql + cnt - 1);
solve(mid + 1, r, ql + cnt, qr);
}

完整代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#define space putchar(' ')
#define enter putchar('\n')
using namespace std;
typedef long long ll;
template <class T>
void read(T &x){
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-') op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}

const int N = 505, M = 60005;
int n, m, tr

, ans[M], idx, sum[M];
struct element {
int x, y, val;
bool operator < (const element &b) const{
return val < b.val;
}
} a[N*N];
struct query {
int id, xa, ya, xb, yb, k;
} q[M], tmp[M];

void add(int x, int y, int val){
for(int px = x; px <= n; px += px & -px)
for(int py = y; py <= n; py += py & -py)
tr[px][py] += val;
}
int single_ask(int x, int y){
int ret = 0;
for(int px = x; px; px -= px & -px)
for(int py = y; py; py -= py & -py)
ret += tr[px][py];
return ret;
}
int ask(query Q){
return single_ask(Q.xb, Q.yb) - single_ask(Q.xa - 1, Q.yb) - single_ask(Q.xb, Q.ya - 1) + single_ask(Q.xa - 1, Q.ya - 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++)
ans[q[i].id] = a[l].val;
return;
}
int mid = (l + r) >> 1, cnt = 0;
for(int i = l; i <= mid; i++)
add(a[i].x, a[i].y, 1);
for(int i = ql; i <= qr; i++){
sum[i] = ask(q[i]);
if(sum[i] >= q[i].k) cnt++;
}
for(int i = ql, pl = ql, pr = ql + cnt; i <= qr; i++)
if(sum[i] >= q[i].k) tmp[pl++] = q[i];
else q[i].k -= sum[i], tmp[pr++] = q[i];
for(int i = ql; i <= qr; i++) q[i] = tmp[i];
for(int i = l; i <= mid; i++)
add(a[i].x, a[i].y, -1);
solve(l, mid, ql, ql + cnt - 1);
solve(mid + 1, r, ql + cnt, qr);
}

int main(){
read(n), read(m);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
a[++idx].x = i, a[idx].y = j, read(a[idx].val);
sort(a + 1, a + idx + 1);
for(int i = 1; i <= m; i++)
q[i].id = i, read(q[i].xa), read(q[i].ya), read(q[i].xb), read(q[i].yb), read(q[i].k);
solve(1, idx, 1, m);
for(int i = 1; i <= m; i++)
write(ans[i]), enter;
return 0;
}