Crowd HDU - 4456 曼哈顿距离转切比雪夫距离的坐标变换与二维坐标离散化

通过坐标变换x'=x-y+n和y'=x+y，开大两倍空间就可以用一个两倍大的二维矩阵中

从[x'-a,y'-a]到[x'+a,y'+a]这个子矩阵表示原图中到(x,y)曼哈顿距离小于等于a的菱形区域啦

二维的离散化就是把(x,y)映射到x*n+y这个一维数列上

这题极限空间应该是log(2n)*log(2n)*m吧。。这是2e7左右了呀，不知道4e6怎么过的？可能离散不出那么多严格的logN吧

//#include<bits/stdc++.h>  
//#pragma comment(linker, "/STACK:1024000000,1024000000")   
#include<stdio.h>  
#include<algorithm>  
#include<queue>  
#include<string.h>  
#include<iostream>  
#include<math.h>                    
#include<stack>
#include<set>  
#include<map>  
#include<vector>  
#include<iomanip> 
#include<bitset>
using namespace std;         //

#define ll long long  
#define ull unsigned long long
#define pb push_back  
#define FOR(a) for(int i=1;i<=a;i++) 
#define sqr(a) (a)*(a)
#define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))
ll qp(ll a,ll b,ll mod){
    ll t=1;while(b){if(b&1)t=t*a%mod;b>>=1;a=a*a%mod;}return t;
}
struct DOT{int x;int y;};
inline void read(int &x){int k=0;char f=1;char c=getchar();for(;!isdigit(c);c=getchar())if(c=='-')f=-1;for(;isdigit(c);c=getchar())k=k*10+c-'0';x=k*f;} 
void ex(){puts("No");exit(0);}
const int dx[4]={0,0,-1,1};
const int dy[4]={1,-1,0,0};
const int inf=0x3f3f3f3f; 
const ll Linf=0x3f3f3f3f3f3f3f3fLL;
const ll Mod=1e18+7;
const double eps=1e-6;
const double pi=acos(-1.0);



inline int lowbit(int x){return x&-x;}

const int maxn=8e4+33;

int op[maxn],posx[maxn],posy[maxn];
int v[maxn],h[4000050],tree[4000050],n,m,N,cnt;

void add(int x,int y,int z){
    for(int i=x;i<=N;i+=lowbit(i)){
        for(int j=y;j<=N;j+=lowbit(j)){
            int pos=lower_bound(h,h+cnt,i*N+j)-h;
            tree[pos]+=z;
        }
    }
}
int ask(int x,int y){
    int ans=0;
    for(int i=x;i>0;i-=lowbit(i)){
        for(int j=y;j>0;j-=lowbit(j)){
            int pos=lower_bound(h,h+cnt,i*N+j)-h;
            if(h[pos]==i*N+j)ans+=tree[pos];
        }
    }
    return ans;
}

int main(){
    while(scanf("%d",&n)&&n){
        memset(tree,0,sizeof tree);
        int x,y;
        N=2*n;
        cnt=0;
        scanf("%d",&m);
        for(int i=1;i<=m;i++){
            scanf("%d%d%d%d",&op[i],&x,&y,&v[i]);
            int xx=x-y+n;int yy=x+y;
            posx[i]=xx;
            posy[i]=yy;
            if(op[i]==1){
                for(int j=xx;j<=N;j+=lowbit(j)){
                    for(int k=yy;k<=N;k+=lowbit(k)){
                        h[cnt++]=j*N+k;
                    }
                }
            }
        }
        sort(h,h+cnt);
        for(int i=1;i<=m;i++){
            if(op[i]==1)add(posx[i],posy[i],v[i]);
            else{
                int X1=max(1,posx[i]-v[i]),Y1=max(1,posy[i]-v[i]);
                int X2=min(N,posx[i]+v[i]),Y2=min(N,posy[i]+v[i]);
                printf("%d\n",ask(X2,Y2)+ask(X1-1,Y1-1)
                        -ask(X2,Y1-1)-ask(X1-1,Y2)
                        );
            }
        }
    }
}