poj 2482 Stars in Your Window

Stars in Your Window
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 12502   Accepted: 3428

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

题意:用一个长方形覆盖平面上尽可能多的点。
思路:线段树+平面扫描,以每个星星为长方形的中心画长方形,这样平面就有n个长方形,这时长方形重叠次数最多的区域就是我要找的中心点所在的区域,换句话说,若所要找的中心点在这块区域,那么以该点为中心的长方形能覆盖尽可能多的星星。
那么如何找到这块区域呢?可以用线段树,维护值为纵坐标每一段区域内的星星总亮度。设星星坐标(x,y),那么记录长方形的两条竖边为(x-W/2,y-H/2,y+H/2)和(x+W/2,y-H/2,y+H/2).当扫描线经过左边那条竖边,那么在线段树的[y-H/2,y+H/2)区域内亮度值加上该星星的亮度c.
当经过右边的竖边时同理,该区域亮度值减c。
AC代码:
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const int N_MAX = 10000 + 4;
ll xs[N_MAX], ys[N_MAX]; int cs[N_MAX];
ll X[2* N_MAX], Y[2 * N_MAX];//记录以每一个点为中心所构成的长方形的左右两竖边位置
int Data[N_MAX*8], dat_max[N_MAX*8];//!!!!
pair<pair<int, int>, pair<int, int> >event[2 * N_MAX];//x<->c,y1<->y2,排序时按照x坐标大小来排
int n, W, H;
//[from,to)区间加上v值,i节点编号,对应区间[l,r)
void add(int from,int to,int v,int i,int l,int r) {
    if (from <= l&&to >= r) {//完全包含i节点控制的区间,全员都要加上v
        Data[i] += v;//如果整个区间都加上v,不用交给儿子处理了
        dat_max[i] += v;
        return;
    }
    int middle = (l+r) >> 1;// 
    if (from < middle) add(from, to,v, 2 * i, l, middle);
    if (to > middle)add(from, to, v, 2 * i + 1, middle, r);
    dat_max[i] = max(dat_max[2 * i], dat_max[2 * i + 1]) + Data[i];
}

int main() {
    while (scanf("%d%d%d",&n,&W,&H)!=EOF) {
        memset(Data, 0, sizeof(Data));
        memset(dat_max, 0, sizeof(dat_max));
        for (int i = 0; i < n;i++) {
            scanf("%lld%lld%d",xs+i,ys+i,cs+i);
            xs[i] *= 2;//后面W和H如果除以2可能会有误差,不如这里先乘2
            ys[i] *= 2;
        }
        for (int i = 0; i < n;i++) {
            X[2 * i] = xs[i] - W;
            X[2 * i + 1] = xs[i] + W;
            Y[2 * i] = ys[i] - H;
            Y[2 * i + 1] = ys[i] + H;
        }
        sort(X, X + 2 * n);
        sort(Y, Y + 2 * n);
        for (int i = 0; i <  n;i++) {//坐标离散化
            int y1 = lower_bound(Y, Y + 2 * n, ys[i] - H) - Y;
            int y2 = lower_bound(Y, Y + 2 * n, ys[i] + H) -Y;
            int x1 = lower_bound(X, X + 2 * n, xs[i] - W) - X;
            int x2 = lower_bound(X, X + 2 * n, xs[i] + W) - X;
            event[2 * i] = make_pair(make_pair(x1, cs[i]), make_pair(y1, y2));
            event[2 * i+1] = make_pair(make_pair(x2, -cs[i]), make_pair(y1, y2));//!!!!
           
        }
        sort(event, event + 2 * n);
        int res = 0;
        for (int i = 0; i < 2 * n;i++) {
            add(event[i].second.first, event[i].second.second, event[i].first.second, 1, 0, 2 * n+1 );
            res = max(res, dat_max[1]);
        }
        printf("%d\n",res);
    }
    return 0;

}

 



posted on 2017-09-04 19:31  ZefengYao  阅读(147)  评论(0编辑  收藏  举报

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