BZOJ 2738 矩阵乘法

整体二分。直接while就可以啊。。。好像复杂度确实是对的。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 250050
using namespace std;
struct pnt
{
    int x,y,val;
}p[maxn];
struct query
{
    int x1,y1,x2,y2,id,k;
}q[maxn];
int n,m,x,t[550][550],cnt=0,ans[maxn],id[maxn],mx=0,T=0,tmp[maxn];
bool vis[maxn];
bool cmp(pnt x,pnt y)
{
    return x.val<y.val;
}
int lowbit(int x) {return (x&(-x));}
void add(int x,int y,int val)
{
    for (int i=x;i<=n;i+=lowbit(i))
        for (int j=y;j<=n;j+=lowbit(j))
            t[i][j]+=val;
}
int ask(int x,int y)
{
    int ret=0;
    for (int i=x;i>=1;i-=lowbit(i))
        for (int j=y;j>=1;j-=lowbit(j))
            ret+=t[i][j];
    return ret;
}
int find(int x)
{
    return ask(q[x].x2,q[x].y2)-ask(q[x].x1-1,q[x].y2)-ask(q[x].x2,q[x].y1-1)+ask(q[x].x1-1,q[x].y1-1);
}
void solve(int l,int r,int L,int R)
{
    if (l==r) return;
    if (L>R) return;
    int mid=l+r>>1,l1,l2,ret=0;
    while ((p[T+1].val<=mid) && (T<cnt)) {add(p[T+1].x,p[T+1].y,1);T++;}
    while ((p[T].val>mid) && (T>=1)) {add(p[T].x,p[T].y,-1);T--;}
    for (int i=L;i<=R;i++)
    {
        int now=find(id[i]);
        if (now>q[id[i]].k-1) {vis[i]=true;ans[id[i]]=mid;ret++;}
        else vis[i]=false;
    }
    l1=L;l2=L+ret;
    for (int i=L;i<=R;i++)
    {
        if (vis[i]) tmp[l1++]=id[i];
        else tmp[l2++]=id[i];
    }
    for (int i=L;i<=R;i++) id[i]=tmp[i];
    solve(l,mid,L,l1-1);solve(mid+1,r,l1,l2-1);
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
        {
            scanf("%d",&x);
            cnt++;mx=max(mx,x);
            p[cnt].x=i;p[cnt].y=j;p[cnt].val=x;
        }
    sort(p+1,p+cnt+1,cmp);
    for (int i=1;i<=m;i++)
    {
        scanf("%d%d%d%d%d",&q[i].x1,&q[i].y1,&q[i].x2,&q[i].y2,&q[i].k);
        q[i].id=i;                                                                                                           
    }
    for (int i=1;i<=m;i++) id[i]=i;
    solve(0,mx+1,1,m);
    for (int i=1;i<=m;i++) printf("%d\n",ans[i]);
    return 0;
}