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
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <cmath> 6 using namespace std; 7 const int maxn = 81111; 8 const int maxm = 3e6+50; //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。 9 int p[maxn],x[maxn],y[maxn],z[maxn]; 10 int arr[maxm]; 11 int cnt[maxm]; 12 int num = 0; 13 int len = 0; 14 int w = 0; 15 int lowbit(int x) 16 { 17 return x&(-x); 18 } 19 void pre(int x,int y) 20 { 21 for(int i=x;i<=w;i+=lowbit(i)) 22 { 23 for(int j=y;j<=w;j+=lowbit(j)) 24 { 25 cnt[num++] = i*w+j; 26 } 27 } 28 } 29 void sum(int x,int y,int tt) 30 { 31 for(int i=x;i<=w;i+=lowbit(i)) 32 { 33 for(int j=y;j<=w;j+=lowbit(j)) 34 { 35 int id = lower_bound(cnt,cnt+len,i*w+j)-cnt; 36 arr[id] += tt; 37 } 38 } 39 } 40 int getsum(int x,int y) 41 { 42 x=max(x,0); x=min(x,w); //边界 43 y=max(y,0); y=min(y,w); 44 int summ = 0; 45 for(int i=x;i>0;i-=lowbit(i)) 46 { 47 for(int j=y;j>0;j-=lowbit(j)) 48 { 49 int id = lower_bound(cnt,cnt+len,i*w+j)-cnt; 50 if(cnt[id]==(i*w+j)) summ += arr[id]; 51 } 52 } 53 return summ; 54 } 55 int main() 56 { 57 int n,m; 58 while(scanf("%d",&n)!=EOF&&n) 59 { 60 scanf("%d",&m); 61 w = 2*n; 62 memset(cnt,0,sizeof(cnt)); 63 memset(arr,0,sizeof(arr)); 64 num = 0; 65 for(int i=1;i<=m;i++) 66 { 67 scanf("%d %d %d %d",&p[i],&x[i],&y[i],&z[i]); 68 int nx = x[i]-y[i]+n; 69 int ny = x[i]+y[i]; 70 if(p[i]==1) 71 pre(nx,ny); 72 } 73 sort(cnt,cnt+num); 74 len = unique(cnt,cnt+num)-cnt; 75 for(int i=1;i<=m;i++) 76 { 77 int nx = x[i]-y[i]+n; 78 int ny = x[i]+y[i]; 79 if(p[i]==1) 80 { 81 sum(nx,ny,z[i]); 82 } 83 else 84 { 85 int x1,y1,x2,y2,x3,y3,x4,y4; 86 x1 = nx+z[i];y1 = ny+z[i]; 87 x2 = nx+z[i];y2 = ny-z[i];y2--; 88 x3 = nx-z[i];y3 = ny+z[i];x3--; 89 x4 = nx-z[i];y4 = ny-z[i];x4--;y4--; 90 printf("%d ",getsum(x1,y1)-getsum(x2,y2)-getsum(x3,y3)+getsum(x4,y4)); 91 } 92 } 93 } 94 return 0; 95 }