給出N個點,和一個w*h的矩形
給出N個點的坐標,求該矩形最多可以覆蓋多少個點
對每個點point(x,y)右邊生成對應的點(x+w,y)值為-1;
縱向建立線段樹,從左到右掃描線掃一遍,遇到點則用該點的權值更新區間(y,y+h)
#include "stdio.h"
#include "string.h"
#include "algorithm"
using namespace std;
struct Mark
{
int x,y,s;
}mark[30010];
struct node
{
int l,r,x,lazy;
}data[200010];
bool cmp(Mark a, Mark b)
{
if (a.x!=b.x)
return a.xb.s;
}
int Max(int a,int b)
{
if (amid) updata(l,r,k*2+1,op);
else
{
updata(l,mid,k*2,op);
updata(mid+1,r,k*2+1,op);
}
data[k].x=Max(data[k*2].x,data[k*2+1].x);
}
int main()
{
int n,w,h,i,x,y,ans;
while (scanf("%d",&n)!=EOF)
{
if (n<0) break;
scanf("%d%d",&w,&h);
for (i=0;i40000) y=40000;
updata(mark[i].y,y,1,mark[i].s);
ans=Max(ans,data[1].x);
}
printf("%d\n",ans);
}
return 0;
}