POJ2482 Stars in Your Window 题解

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.

扫描线学得人好迷啊……虽然题目背景挺好的……

边界问题把我整佛了，草……

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对于本题的思维转化，网上的题解一大把，在这方面已经说的非常清楚了。这篇文章只讨论两点：第一，扫描线的本质是什么；第二，边界问题的处理。

首先第一点，扫描线一般用来解决二维图形的交并问题，其实质是用一条边的顺序遍历将其中一维有序，另一维便可用区间数据结构维护。在本题中，我们把扫描线设为一条竖直直线使得水平方向有序，竖直方向的线段便可用线段树维护。可以将竖直方向切割出来的各个线段想象为一个区间，线段树的一个节点便维护了一条被切割的线段（即一个节点的 \(l,r\) 就对应原图形中被切割线段的左右端点）。

第二点，本题的边界问题。考虑到矩形的边界无法覆盖，而又考虑到星星的坐标均为整数，便稍作变换，将星星的横纵坐标均减去 \(0.5\) 个单位长度，变为 \((x-0.5,y-0.5)\)。此时再假设矩形的坐标均为整数，那么此时矩形的对角线的两点变为 \((x,y)\) 与 \((x+w-1,y+h-1)\)，这么做的目的是让矩形的边界也包含在内，便于计算。一般讲到这，基本上所有的题解便开始放代码了，无非就是再多说明一句排序时的注意事项，但是根本不解释为什么要这么排序。

考虑这样一张图：

红方块代表星星（平移 \(0.5\) 没有画出来），其范围已用三种不同颜色标识出来，橙色是扫描线。可以看到，如果按照排序时先算出边再算入边的方法，那么就会出现误差，而这误差就是体现在原图的深蓝色区域。这是坐标平移所带来的误差，所以补救措施就是修改排序使其先算入边再算出边。

#include <cstdio>
#include <algorithm>
using namespace std;

const int N=10000;
struct Segment
{
    int x,y1,y2,c;
    bool operator<(const Segment &a) {return x==a.x?c>a.c:x<a.x;}
}Seg[(N<<1)+5];
struct Tree
{
    int l,r,sum,tag;
    #define l(x) t[x].l
    #define r(x) t[x].r
}t[(N<<3)+5];
int n,w,h,lsy[(N<<1)+5];

void build(int rt,int l,int r)
{
    l(rt)=l,r(rt)=r;
    t[rt].sum=t[rt].tag=0;
    if(l==r) return;
    int mid=l+r>>1;
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
}

inline void pushdown(int rt)
{
    t[rt<<1].tag+=t[rt].tag;
    t[rt<<1|1].tag+=t[rt].tag;
    t[rt<<1].sum+=t[rt].tag;
    t[rt<<1|1].sum+=t[rt].tag;
    t[rt].tag=0;
}

void upd(int rt,int l,int r,int k)
{
    if(l<=l(rt)&&r>=r(rt))
        {t[rt].sum+=k,t[rt].tag+=k; return;}
    if(t[rt].tag) pushdown(rt);
    int mid=l(rt)+r(rt)>>1;
    if(l<=mid) upd(rt<<1,l,r,k);
    if(r>mid) upd(rt<<1|1,l,r,k);
    t[rt].sum=max(t[rt<<1].sum,t[rt<<1|1].sum);
}

int main()
{
    while(~scanf("%d%d%d",&n,&w,&h))
    {
        int ans=0;
        for(int i=1;i<=n;++i)
        {
            int x,y,c;
            scanf("%d%d%d",&x,&y,&c);
            Seg[2*i-1]=(Segment){x,y,y+h-1,c};
            Seg[2*i]=(Segment){x+w-1,y,y+h-1,-c};
            lsy[2*i-1]=y,lsy[2*i]=y+h-1;
        }
        n<<=1;
        sort(lsy+1,lsy+n+1);
        int cnt=unique(lsy+1,lsy+n+1)-lsy-1;
        build(1,1,cnt);
        sort(Seg+1,Seg+n+1);
        for(int i=1;i<=n;++i)
        {
            int l=lower_bound(lsy+1,lsy+cnt+1,Seg[i].y1)-lsy;
            int r=lower_bound(lsy+1,lsy+cnt+1,Seg[i].y2)-lsy;
            upd(1,l,r,Seg[i].c);
            ans=max(ans,t[1].sum);
        }
        printf("%d\n",ans);
    }
    return 0;
}