[poj P2482] Stars in Your Window

[poj P2482] Stars in Your Window

Time Limit: 1000MS  Memory Limit: 65536K

Description

Fleeting time does not blur my memory of you. Can it really be 4 years since I first saw you? I still remember, vividly, on the beautiful Zhuhai Campus, 4 years ago, from the moment I saw you smile, as you were walking out of the classroom and turned your head back, with the soft sunset glow shining on your rosy cheek, I knew, I knew that I was already drunk on you. Then, after several months’ observation and prying, your grace and your wisdom, your attitude to life and your aspiration for future were all strongly impressed on my memory. You were the glamorous and sunny girl whom I always dream of to share the rest of my life with. Alas, actually you were far beyond my wildest dreams and I had no idea about how to bridge that gulf between you and me. So I schemed nothing but to wait, to wait for an appropriate opportunity. Till now — the arrival of graduation, I realize I am such an idiot that one should create the opportunity and seize it instead of just waiting.

These days, having parted with friends, roommates and classmates one after another, I still cannot believe the fact that after waving hands, these familiar faces will soon vanish from our life and become no more than a memory. I will move out from school tomorrow. And you are planning to fly far far away, to pursue your future and fulfill your dreams. Perhaps we will not meet each other any more if without fate and luck. So tonight, I was wandering around your dormitory building hoping to meet you there by chance. But contradictorily, your appearance must quicken my heartbeat and my clumsy tongue might be not able to belch out a word. I cannot remember how many times I have passed your dormitory building both in Zhuhai and Guangzhou, and each time aspired to see you appear in the balcony or your silhouette that cast on the window. I cannot remember how many times this idea comes to my mind: call her out to have dinner or at least a conversation. But each time, thinking of your excellence and my commonness, the predominance of timidity over courage drove me leave silently.

Graduation, means the end of life in university, the end of these glorious, romantic years. Your lovely smile which is my original incentive to work hard and this unrequited love will be both sealed as a memory in the deep of my heart and my mind. Graduation, also means a start of new life, a footprint on the way to bright prospect. I truly hope you will be happy everyday abroad and everything goes well. Meanwhile, I will try to get out from puerility and become more sophisticated. To pursue my own love and happiness here in reality will be my ideal I never desert.

Farewell, my princess!

If someday, somewhere, we have a chance to gather, even as gray-haired man and woman, at that time, I hope we can be good friends to share this memory proudly to relight the youthful and joyful emotions. If this chance never comes, I wish I were the stars in the sky and twinkling in your window, to bless you far away, as friends, to accompany you every night, sharing the sweet dreams or going through the nightmares together. 

Here comes the problem: Assume the sky is a flat plane. All the stars lie on it with a location (x, y). for each star, there is a grade ranging from 1 to 100, representing its brightness, where 100 is the brightest and 1 is the weakest. The window is a rectangle whose edges are parallel to the x-axis or y-axis. Your task is to tell where I should put the window in order to maximize the sum of the brightness of the stars within the window. Note, the stars which are right on the edge of the window does not count. The window can be translated but rotation is not allowed. 

Input

There are several test cases in the input. The first line of each case contains 3 integers: n, W, H, indicating the number of stars, the horizontal length and the vertical height of the rectangle-shaped window. Then n lines follow, with 3 integers each: x, y, c, telling the location (x, y) and the brightness of each star. No two stars are on the same point.

There are at least 1 and at most 10000 stars in the sky. 1<=W，H<=1000000, 0<=x，y<2^31.

Output

For each test case, output the maximum brightness in a single line.

Sample Input

3 5 4
1 2 3
2 3 2
6 3 1
3 5 4
1 2 3
2 3 2
5 3 1

Sample Output

5
6

Source

POJ Contest,Author:kinfkong@ZSU

这封情书。。。写的可真有水平啊。。值得借鉴qwq

好吧，这其实就是非常经典的扫描线题。

这种题目，我们先需要转换一下。

题目中用矩形框点，可以等价变换为求以每个点为角（任意一个）的矩形的最大交。

这样，我们相当于把原问题变成了求出矩形面积并，只是另类一点而已（还简单了不少）。

那么，我们可以用一个差分的思想，将一个点(x,y,z)的信息先转为一个矩形(x,y,W,H,z)，再变成两条线段。

这两条线段的信息分别为(y,x,x+W,z)和(y+H,x,x+W,-z)。

其中值得注意的是，题目中说了，不包含边界上面的点，所以上方的两条线段实际上只包含左端点而不包含右端点。

然后这就变成了加权线段覆盖问题。完全就是线段树的范围。

但是我们还要先按x坐标离散一下，再建立，更新线段树。

当然还要注意，我们更新线段树要有顺序，比如从上往下或从下往上都可以。

但还要注意的是，只有当你把同一高度的线段都处理完了，才能进行查询，更新答案，否则可能会导致负标记还没删完，答案就更新了。

code：

 %:pragma GCC optimize()
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define LL long long
 using namespace std;
 ;
 int n,cnt,ans; LL a[N],W,H;
 class node {
     private:
         int v,t; node *l,*r;
     public:
         #define m ((l)+(r)>>1)
         node() {v=t=,l=r=NULL;}
         inline void pushup(node* &cu) {
             cu->v=;
             ) cu->v=max(cu->v,cu->l->v);
             ) cu->v=max(cu->v,cu->r->v);
         }
         inline void pushdown(node* &cu) {
             ) cu->l->v+=cu->t,cu->l->t+=cu->t;
             ) cu->r->v+=cu->t,cu->r->t+=cu->t;
             cu->t=;
         }
         inline void setup(node* &cu,int l,int r) {
             cu=new node;
             ; return;}
             setup(cu->l,l,m),setup(cu->r,m+,r);
             pushup(cu);
         }
         inline void update(node* &cu,int l,int r,int aiml,int aimr,int v) {
             if (aiml>r||aimr<l) return;
             if (l>=aiml&&r<=aimr) {cu->v+=v; cu->t+=v; return;}
             pushdown(cu);
             if (aimr<=m) update(cu->l,l,m,aiml,aimr,v); else
             ,r,aiml,aimr,v); else
             update(cu->l,l,m,aiml,aimr,v),update(cu->r,m+,r,aiml,aimr,v);
             pushup(cu);
         }
         inline int answer(node* &cu) {return cu->v;}
 }t,*root;
 struct line {
     LL x,l,r; int z;
     line () {}
     line (LL _x,LL _l,LL _r,int _z):x(_x),l(_l),r(_r),z(_z) {}
 }s[N];
 inline bool cmp(const line &u,const line &v) {return u.x<v.x;}
 inline int read() {
     ; char ch=getchar();
     ') ch=getchar();
     +ch-',ch=getchar();
     return x;
 }
 inline ,root=;}
 int main() {
     while (scanf("%d%lld%lld",&n,&W,&H)!=EOF) {
         init();
         ; i<=n; i++) {
             LL x=read(),y=read(); int z=read();
             s[(i-)<<|]=line(y,x,x+W,z);
             s[i<<]=line(y+H,x,x+W,-z);
             a[(i-)<<|]=x,a[i<<]=x+W;
         }
         sort(a+,a++n+n);
         cnt=unique(a+,a++n+n)-a-;
         sort(s+,s++n+n,cmp);
         t.setup(root,,cnt);
         ,l,r; i<=n<<; i++) {
             ].x&&i>) ans=max(ans,t.answer(root));
             l=lower_bound(a+,a+cnt+,s[i].l)-a;
             r=lower_bound(a+,a+cnt+,s[i].r)-a-;
             t.update(root,,cnt,l,r,s[i].z);
         }
         printf("%d\n",ans);
     }
     ;
 }