HDU 4456（二维树状数组+坐标转换）

题目链接：Problem - 4456

看别人叙述看的心烦，于是我自己画了一张图。

上图。

上代码

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;
 const int maxn = ;
 const int maxm = 3e6+; //这里大小的确定应该是80000*log2(20000)*log2(20000),算出来大约是一千三百万，数据应该是水一些，然而看所有人的题解，都是四百万，呵呵呵，这其中一定有什么交易。
 int p[maxn],x[maxn],y[maxn],z[maxn];
 int arr[maxm];
 int cnt[maxm];
 int num = ;
 int len = ;
 int w = ;
 int lowbit(int x)
 {
     return x&(-x);
 }
 void pre(int x,int y)
 {
     for(int i=x;i<=w;i+=lowbit(i))
     {
         for(int j=y;j<=w;j+=lowbit(j))
         {
             cnt[num++] = i*w+j;
         }
     }
 }
 void sum(int x,int y,int tt)
 {
     for(int i=x;i<=w;i+=lowbit(i))
     {
         for(int j=y;j<=w;j+=lowbit(j))
         {
             int id = lower_bound(cnt,cnt+len,i*w+j)-cnt;
             arr[id] += tt;
         }
     }
 }
 int getsum(int x,int y)
 {
     x=max(x,); x=min(x,w); //边界
     y=max(y,); y=min(y,w);
     int summ = ;
     for(int i=x;i>;i-=lowbit(i))
     {
         for(int j=y;j>;j-=lowbit(j))
         {
             int id = lower_bound(cnt,cnt+len,i*w+j)-cnt;
             if(cnt[id]==(i*w+j)) summ += arr[id];
         }
     }
     return summ;
 }
 int main()
 {
     int n,m;
     while(scanf("%d",&n)!=EOF&&n)
     {
         scanf("%d",&m);
         w = *n;
         memset(cnt,,sizeof(cnt));
         memset(arr,,sizeof(arr));
         num = ;
         for(int i=;i<=m;i++)
         {
             scanf("%d %d %d %d",&p[i],&x[i],&y[i],&z[i]);
             int nx = x[i]-y[i]+n;
             int ny = x[i]+y[i];
             if(p[i]==)
             pre(nx,ny);
         }
         sort(cnt,cnt+num);
         len = unique(cnt,cnt+num)-cnt;
         for(int i=;i<=m;i++)
         {
             int nx = x[i]-y[i]+n;
             int ny = x[i]+y[i];
             if(p[i]==)
             {
                 sum(nx,ny,z[i]);
             }
             else
             {
                 int x1,y1,x2,y2,x3,y3,x4,y4;
                 x1 = nx+z[i];y1 = ny+z[i];
                 x2 = nx+z[i];y2 = ny-z[i];y2--;
                 x3 = nx-z[i];y3 = ny+z[i];x3--;
                 x4 = nx-z[i];y4 = ny-z[i];x4--;y4--;
                 printf("%d\n",getsum(x1,y1)-getsum(x2,y2)-getsum(x3,y3)+getsum(x4,y4));
             }
         }
     }
     return ;
 }