HDU

题意:给一个矩阵,两种操作1:修改单点的权值,2:查询和某个点曼哈顿距离小于r点的权值和
题解:先旋转坐标轴,(x,y)->(x-y,x+y)然后就变成了cdq分治裸题,子矩阵和和单点修改一维时间,二维xcdq,三维ybit

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("c.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=320000+10,maxn=200000+10,inf=0x3f3f3f3f;

struct query{int op,x,y,z,id;}p[N];
bool cmp(const query&a,const query&b){return a.x<b.x || (a.x==b.x&&a.y<b.y);}
int n,m,cnt,ans[N],res;
struct bit{
    int val[N];
    void update(int i,int v)
    {
        for(;i<N;i+=i&(-i))val[i]+=v;
    }
    int query(int i)
    {
        int ans=0;
        for(;i;i-=i&(-i))ans+=val[i];
        return ans;
    }
}b;
void cdq(int l,int r)
{
    if(l==r)return ;
    int mid=(l+r)>>1;
    cdq(l,mid);cdq(mid+1,r);
    sort(p+l,p+mid+1,cmp);sort(p+mid+1,p+r+1,cmp);
    int le=l,ri=mid+1;
    while(ri<=r)
    {
        while(le<=mid&&p[le].op==2)le++;
        while(ri<=r&&p[ri].op==1)ri++;
        if(ri > r) break;
        if(le<=mid&&p[le].x<=p[ri].x)b.update(p[le].y,p[le].z),le++;
        else ans[p[ri].id]+=p[ri].z*b.query(p[ri].y),ri++;
    }
    le--;
    for(;le>=l;le--)if(p[le].op==1)b.update(p[le].y,-p[le].z);
}
int main()
{
    while(~scanf("%d",&n))
    {
        if(!n)break;
        scanf("%d",&m);
        cnt=res=0;
        for(int i=1;i<=m;i++)
        {
            ++cnt;
            scanf("%d%d%d%d",&p[cnt].op,&p[cnt].x,&p[cnt].y,&p[cnt].z);
            p[cnt].id=0;
            int x=p[cnt].x,y=p[cnt].y,z=p[cnt].z;
            p[cnt].x=x-y+n,p[cnt].y=x+y;
            x=p[cnt].x,y=p[cnt].y;
            if(p[cnt].op==2)
            {
                p[cnt].x=MIN(x+z,2*n + 5);
                p[cnt].y=MIN(y+z,2*n + 5);
                p[cnt].z=1;
                p[cnt].id=++res;
                ans[res] = 0;
                if(x-z-1>0)
                {
                    ++cnt;p[cnt].id=res;p[cnt].op=2;
                    p[cnt].x=x-z-1;p[cnt].y=y+z;p[cnt].z=-1;
                }
                if(y-z-1>0)
                {
                    ++cnt;p[cnt].id=res;p[cnt].op=2;
                    p[cnt].x=x+z;p[cnt].y=y-z-1;p[cnt].z=-1;
                }
                if(x-z-1>0&&y-z-1>0)
                {
                    ++cnt;p[cnt].id=res;p[cnt].op=2;
                    p[cnt].x=x-z-1;p[cnt].y=y-z-1;p[cnt].z=1;
                }
            }
        }
//        for(int i=1;i<=cnt;i++)printf("%d %d %d %d %d
",p[i].op,p[i].x,p[i].y,p[i].z,p[i].id);
        cdq(1,cnt);
        for(int i=1;i<=res;i++)printf("%d
",ans[i]);
    }
    return 0;
}
/********************
8 5
1 8 8 1
1 1 1 -2
2 5 5 6
1 5 5 3
2 2 3 9
********************/