题意：给出一个h*w的矩形，再给出n个坐标，在这n个坐标种树，再给出一个s*t大小的矩形，问在这个s*t的矩形里面最多能够得到多少棵树
二维的树状数组，求最多能够得到的树的时候，因为h,w都不超过500，直接暴力
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include <cmath> 5 #include<stack> 6 #include<vector> 7 #include<map> 8 #include<set> 9 #include<queue> 10 #include<algorithm> 11 using namespace std; 12 13 typedef long long LL; 14 const int INF = (1<<30)-1; 15 const int mod=1000000007; 16 const int maxn=1005; 17 18 int c[maxn][maxn]; 19 20 int lowbit(int x){ return x & (-x);} 21 22 int sum(int x,int y){ 23 int ret=0; 24 int y1; 25 while(x > 0){ 26 y1=y; 27 while(y1 > 0){ 28 ret += c[x][y1];y1-=lowbit(y1); 29 } 30 x-=lowbit(x); 31 } 32 return ret; 33 } 34 35 void add(int x,int y,int d){ 36 int y1; 37 while(x < maxn){ 38 y1=y; 39 while(y1 < maxn){ 40 c[x][y1] += d;y1+=lowbit(y1); 41 } 42 x+=lowbit(x); 43 } 44 } 45 46 int main(){ 47 int n; 48 while(scanf("%d",&n)!=EOF && n){ 49 memset(c,0,sizeof(c)); 50 int w,h; 51 scanf("%d %d",&w,&h); 52 while(n--){ 53 int u,v; 54 scanf("%d %d",&u,&v); 55 add(v,u,1); 56 } 57 int x,y; 58 scanf("%d %d",&x,&y);swap(x,y); 59 60 int ans=-1; 61 int tmp=0; 62 for(int i=1;i <=h;i++){ 63 for(int j=1;j<=w;j++){ 64 tmp = sum(i+x-1,j+y-1)-sum(i-1,j+y-1)-sum(i+x-1,j-1)+sum(i-1,j-1); 65 ans = max(ans,tmp); 66 } 67 } 68 printf("%d ",ans); 69 } 70 return 0; 71 }