With heavy problems like POJ
This is the question one scan line.
Due to the size of the window is known, we might change my ideas, the problem is transformed into this: There are a number of rectangles (the size is the size of the window position, the lower left corner of the rectangle is a certain position of the stars), each covering a rectangle in a plane the zone has a weight value (luminance stars), seeking a position such that the right position is covered by the maximum value, i.e., the maximum answer.
Why this is right? The size of each window are fixed, if we put on the top right of this window just said rectangular stars, then this star must be seen. By analogy, suppose we have placed in the top right corner of the window as much as possible rectangle range (here that is not appropriate, it should be placed within the weights add up to the greatest extent possible), we will answer as much as possible excellent, so we only need to find the maximum scan line weights can be.
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 const int N=10010; 7 int T; 8 struct line { 9 int x,y1,y2,v; 10 bool operator <(const line &xx) const { 11 if(x==xx.x) return v>xx.v; 12 return x<xx.x; 13 } 14 }a[N<<1]; 15 int n,val[N<<1][2],raw[N<<1]; 16 struct node { 17 int l,r; 18 int data,tag; 19 }s[N<<3]; 20 int ans,t,w,h; 21 inline void build(int now,int l,int r) { 22 s[now].l=l; 23 s[now].r=r; 24 s[now].data=s[now].tag=0; 25 if(l==r) return ; 26 int mid=l+r>>1; 27 build(now<<1,l,mid); 28 build(now<<1|1,mid+1,r); 29 } 30 inline void pushdown(int now) { 31 if(s[now].tag) { 32 int v=s[now].tag; 33 s[now<<1].tag+=v; 34 s[now<<1|1].tag+=v; 35 s[now<<1].data+=v; 36 s[now<<1|1].data+=v; 37 s[now].tag=0; 38 } 39 } 40 void updata(int now,int l,int r,int v) { 41 if(l<=s[now].l&&s[now].r<=r) { 42 s[now].data+=v; 43 s[now].tag+=v; 44 return ; 45 } 46 pushdown(now); 47 int mid=s[now].l+s[now].r>>1; 48 if(l<=mid) updata(now<<1,l,r,v); 49 if(r>mid) updata(now<<1|1,l,r,v); 50 s[now].data=max(s[now<<1].data,s[now<<1|1].data); 51 } 52 int main() { 53 scanf("%d",&T); 54 while(T--) { 55 ans=t=0; 56 scanf("%d%d%d",&n,&w,&h); 57 for(int i=1;i<=n;i++) { 58 int x,y,z; 59 scanf("%d%d%d",&x,&y,&z); 60 a[2*i-1]=(line){x,y,y+h,z}; 61 a[2*i]=(line){x+w-1,y,y+h,-z}; 62 raw[++t]=y; 63 raw[++t]=y+h-1; 64 } 65 sort(a+1,a+2*n+1); 66 sort(raw+1,raw+t+1); 67 t=unique(raw+1,raw+t+1)-raw-1; 68 for(int i=1;i<=2*n;i++) { 69 val[i][0]=lower_bound(raw+1,raw+t+1,a[i].y1)-raw; 70 val[i][1]=lower_bound(raw+1,raw+t+1,a[i].y2)-raw; 71 } 72 build(1,1,t); 73 for(int i=1;i<=2*n;i++) { 74 updata(1,val[i][0],val[i][1]-1,a[i].v); 75 ans=max(ans,s[1].data); 76 } 77 printf("%d\n",ans); 78 } 79 return 0; 80 }