HDU_4456_二维树状数组

http://acm.hdu.edu.cn/showproblem.php?pid=4456

第一道二维树状数组就这么麻烦，题目要计算的是一个菱形范围内的和，于是可以把原来的坐标系旋转45度，就是求一个正方形范围内的和，这里还涉及到坐标系平移和放大，由于题目数据较大，用了离散化来保存需要处理的点，放在h数组内，对应的二维树状数组存在tree数组内。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

inline int lowbit(int x)
{
    return x & (-x);
}
int p[80050],posx[80050],posy[80050],v[80050],h[4500000],tree[4500000],n,m,N,cnt;

void update(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 getsum(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",&p[i],&x,&y,&v[i]);
            int xx = x-y+n,yy = x+y;
            posx[i] = xx;
            posy[i] = yy;
            if(p[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(p[i] == 1)    update(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
",getsum(x2,y2)-getsum(x2,y1-1)-getsum(x1-1,y2)+getsum(x1-1,y1-1));
            }
        }
    }
    return 0;
}