[Luogu1527][国家集训队]矩阵乘法

luogu

题意

给你一个\(N*N\)的矩阵，每次询问一个子矩形的第K小数。（居然连修改都不带的）
\(N\le500,Q\le60000\)

sol

整体二分+二维树状数组裸题。
复杂度是\(O((n^2+Q)\log n^2 \log^2 n)\)也就是\(3\)个\(\log\)吧。

code

#include<cstdio>
#include<algorithm>
using namespace std;
int gi()
{
	int x=0,w=1;char ch=getchar();
	while ((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if (ch=='-') w=0,ch=getchar();
	while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return w?x:-x;
}
const int N = 4e5+5;
struct node{int id,x1,y1,x2,y2,k;}q[N],q1[N],q2[N];
int n,m,cnt,o[N],len,c[505][505],ans[N];
void modify(int x,int y,int v)
{
	for (int i=x;i<=n;i+=i&-i)
		for (int j=y;j<=n;j+=j&-j)
			c[i][j]+=v;
}
int query(int x,int y)
{
	int res=0;
	for (int i=x;i;i-=i&-i)
		for (int j=y;j;j-=j&-j)
			res+=c[i][j];
	return res;
}
void solve(int L,int R,int l,int r)
{
	if (L>R) return;
	if (l==r)
	{
		for (int i=L;i<=R;++i) ans[q[i].id]=l;
		return;
	}
	int mid=l+r>>1,t1=0,t2=0;
	for (int i=L;i<=R;++i)
		if (!q[i].id)
		{
			if (q[i].k<=mid) modify(q[i].x1,q[i].y1,1),q1[++t1]=q[i];
			else q2[++t2]=q[i];
		}
		else
		{
			int tmp=query(q[i].x2,q[i].y2)-query(q[i].x1-1,q[i].y2)-query(q[i].x2,q[i].y1-1)+query(q[i].x1-1,q[i].y1-1);
			if (q[i].k<=tmp) q1[++t1]=q[i];
			else q[i].k-=tmp,q2[++t2]=q[i];
		}
	for (int i=L;i<=R;++i)
		if (!q[i].id&&q[i].k<=mid) modify(q[i].x1,q[i].y1,-1);
	for (int i=L,j=1;j<=t1;++i,++j) q[i]=q1[j];
	for (int i=L+t1,j=1;j<=t2;++i,++j) q[i]=q2[j];
	solve(L,L+t1-1,l,mid);solve(L+t1,R,mid+1,r);
}
int main()
{
	n=gi();m=gi();
	for (int i=1;i<=n;++i)
		for (int j=1;j<=n;++j)
			q[++cnt]=(node){0,i,j,0,0,o[++len]=gi()};
	sort(o+1,o+len+1);len=unique(o+1,o+len+1)-o-1;
	for (int i=1;i<=cnt;++i)
		q[i].k=lower_bound(o+1,o+len+1,q[i].k)-o;
	for (int i=1;i<=m;++i)
		q[++cnt]=(node){i,gi(),gi(),gi(),gi(),gi()};
	solve(1,cnt,1,len);
	for (int i=1;i<=m;++i) printf("%d\n",o[ans[i]]);
	return 0;
}