[题解] poj 2482 Stars in Your Window(线段树+扫描线+面积交)
- 传送门 -
http://poj.org/problem?id=2482
| Time Limit: 1000MS | | Memory Limit: 65536K |
| Total Submissions: 12409 | | Accepted: 3407 |
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 -
某平面上有若干确定坐标的星星(坐标为整数), 每个星星有一个固定的闪亮值, 现有一个固定大小的矩形, 让你用其套出总闪亮值最大的一些星星(边上的不算, 但该矩形的坐标不一定是整点).
- 思路 -
对于每个星星, 我们构造一个以其坐标为左下角的标准形状的矩形, 则以该矩形内的任意一点(边上不算)为右上角构建一个标准矩形都会包含这颗星星.
我们可以看成在这些矩形内部取点(做答案矩形的右上角)都是有收益的, 矩形重合的区域可以获得所有覆盖了该区域的矩形的收益, 这样就很像矩形面积并了, 只是把记录某区间覆盖矩形的数量改为记录某区间的收益(也就是说, 加入的边是有权值的).
于是我们就只要维护一下区间最大值了.
细节见代码.
PS :
1. 附上一丢丢度娘翻译的love letter：
"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. "
"你那可爱的微笑，是我努力工作的原动力，这一份无报答的爱，将永远封存在我的内心深处和我的脑海里。"
"Farewell, my princess! "
"再见，我的公主！"
"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."
"如果这个机会永远不会到来，我希望我是天上的星星，在你的窗前闪烁，愿远方的朋友为你祝福，陪你夜夜，分享甜蜜的梦，一起做噩梦。"
- 代码 -
#include<cstdio>
#include<algorithm>
#define ls rt<<1
#define rs rt<<1|1
using namespace std;
typedef long long ll;
const int M = 1e4 + 5;
struct line {
ll x1, x2, y;
ll v;
}s[M<<1];
struct segtree {
ll val, mx;
}tr[M << 3];
ll pos[M << 1];
ll X, Y;
int n;
bool cmp(line a, line b) {
if (a.y == b.y) return a.v < b.v;
return a.y < b.y;
}
ll Max(ll a, ll b) { return a > b ? a : b; }
void Pushdown(int rt) {
if (tr[rt].val) {
tr[ls].val += tr[rt].val;
tr[rs].val += tr[rt].val;
tr[ls].mx += tr[rt].val;
tr[rs].mx += tr[rt].val;
tr[rt].val = tr[rt].mx = 0;
}
}
void Pushup(int rt) {
tr[rt].mx = Max(tr[ls].mx, tr[rs].mx);
}
void Update(int rt, int l, int r, ll L, ll R, ll v) {
if (L <= pos[l] && pos[r] < R) {
tr[rt].val += v;
tr[rt].mx += v;
}
else {
Pushdown(rt);
int mid = l + r >> 1;
if (L <= pos[mid]) Update(ls, l, mid, L, R, v);
if (pos[mid+1] < R) Update(rs, mid + 1, r, L, R, v);
Pushup(rt);
}
}
int main() {
while (scanf("%d%lld%lld", &n, &X, &Y) != EOF) {
ll x, y, v;
int t = 0;
for (int i = 1; i <= n; ++i) {
scanf("%lld%lld%lld", &x, &y, &v);
s[t].x1 = x, s[t].x2 = x + X, s[t].y = y, s[t].v = v;
pos[t++] = x;
s[t].x1 = x, s[t].x2 = x + X, s[t].y = y + Y, s[t].v = -v;
pos[t++] = x + X;
}
sort(s, s + t, cmp);
sort(pos, pos + t);
int sz = 1;
for (int i = 1; i < t; ++i)
if (pos[i] != pos[i-1])
pos[sz++] = pos[i];
ll ans = 0;
for (int i = 0; i < t; ++i) {
Update(1, 0, sz - 1, s[i].x1, s[i].x2, s[i].v);
ans = Max(ans, tr[1].mx);
}
printf("%lld\n", ans);
}
return 0;
}
