洛谷$P1527$ [国家集训队]矩阵乘法 整体二分

正解:整体二分

解题报告:

传送门$QwQ$

阿看到这种查询若干次第$k$小显然就想到整体二分$QwQ$?

然后现在就只要考虑怎么快速求出一个矩形内所有小于某个数的数的个数?

开始我的想法是离散化然后开个桶前缀和下,然后发现还要计入坐标,显然会$MLE$,就$GG$了$QwQ$

这时候考虑如果是一维,就可以直接树状数组,关于范围的限制可以提前对询问排个序就成.

然后现在是二维?所以用二维树状数组就成昂

然后就做完了?

$QwQ$

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define ri register int
#define rb register bool
#define rc register char
#define lowbit(x) (x&(-x))
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)
#define lb(x) lower_bound(st+1,st+1+sum,x)-st

const int N=6e5+,M=+,inf=1e9;
int n,m,st[N],nod_cnt,tr[M][M],sum;
struct node{int l1,r1,l2,r2,K,id,as;}nod[N],tmp1[N],tmp2[N];

il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch>'' || ch<''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}
il bool cmp(node gd,node gs){return gd.id<gs.id;}
il void modify(ri x,ri y,ri val)
{ri tmp=y;while(x<=n){y=tmp;while(y<=n)tr[x][y]+=val,y+=lowbit(y);x+=lowbit(x);}}
il int query(ri x,ri y)
{ri ret=,tmp=y;while(x){y=tmp;while(y)ret+=tr[x][y],y-=lowbit(y);x-=lowbit(x);}return ret;}
void solv(ri l,ri r,ri dat_l,ri dat_r)
{
    if(l>r)return;
    if(dat_l==dat_r){rp(i,l,r)nod[i].as=dat_l;return;}
    ri mid=(dat_l+dat_r)>>,t1=,t2=;
    rp(i,l,r)
    {
        if(!nod[i].id)if(nod[i].K<=mid)tmp1[++t1]=nod[i],modify(nod[i].l2,nod[i].r2,);else tmp2[++t2]=nod[i];
        else
        {
            ri d=query(nod[i].l2,nod[i].r2)-query(nod[i].l1-,nod[i].r2)-query(nod[i].l2,nod[i].r1-)+query(nod[i].l1-,nod[i].r1-);
            if(d>=nod[i].K)tmp1[++t1]=nod[i];else nod[i].K-=d,tmp2[++t2]=nod[i];
        }
    }
    rp(i,l,r)if(!nod[i].id && nod[i].K<=mid)modify(nod[i].l2,nod[i].r2,-);
    rp(i,,t1)nod[i+l-]=tmp1[i];rp(i,t1+,t1+t2)nod[i+l-]=tmp2[i-t1];
    solv(l,l+t1-,dat_l,mid);solv(l+t1,r,mid+,dat_r);
}

int main()
{
    freopen("1527.in","r",stdin);freopen("1527.out","w",stdout);
    n=read();m=read();rp(i,,n)rp(j,,n)++nod_cnt,nod[nod_cnt]=(node){,,i,j,st[nod_cnt]=read(),};
    sort(st+,st++nod_cnt);sum=unique(st+,st++nod_cnt)-st-;rp(i,,nod_cnt)nod[i].K=lb(nod[i].K);
    rp(i,,m){ri d1=read(),d2=read(),d3=read(),d4=read(),K=read();nod[++nod_cnt]=(node){d1,d2,d3,d4,K,nod_cnt};}
    solv(,nod_cnt,,sum);
    sort(nod+,nod+nod_cnt+,cmp);rp(i,,nod_cnt)if(nod[i].l1)printf("%d\n",st[nod[i].as]);
    return ;
}