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整点上有N颗星星，每颗星星有一个亮度。用一个平行于x轴和y轴，宽为W高为H的方框去套星星。套住的所有星星的亮度之和为S（包括边框上的星星），求S的最大值。
Input
第1行：共3个数N, W, H，中间用空格分割，N为星星的数量，W为方框的宽度，H为方框的高度。(2 <= N <= 50000, 1 <= W, H <= 10^9) 第2 - N + 1行：每行3个数，X, Y, L，中间用空格分隔，分别表示星星的横坐标X，纵坐标Y，以及星星的亮度L。（1 <= X, Y <= 10^9，1 <= L <= 10000)
Output
输出方框能够套住的最大亮度和S。
Input示例
6 3 3 1 1 2 2 2 3 3 3 4 4 4 3 5 5 2 6 6 1
Output示例
12
最近学习线段树，觉得线段树的代码真的是好长啊~，另外线段树的应用范围是真广啊~
能A掉这道题特别开心，这个题目要直接套星星的话不知道怎么套，所以把星星转化成矩形，把矩形转化成小空间或者是线段。然后求哪一个空间，或者那一条线段上的权值最大。
把点转换成矩形就是说把每一个星星的x+xx/2,y+yy/2，y-yy/2。然后就是扫描线，看哪一部分重合的权值最大。
与之前做过的扫描线不同的就是，因为要算上边缘上的星星，所以线段树上的节点不能是某一段空间了，而是具体的一个一个直线了，所以建立线段树的时候就是dycount-1,而不是dycouut-1-1。
然后就是求动态插入某一段线段树，求总的最大值。
代码：
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#include <map>
#pragma warning(disable:4996)
using namespace std;
#define maxn 90002
struct no
{
int le,re;
long long max_value,sing_value;
}tree[maxn*5];
struct li
{
double x,y1,y2;
int value;
bool end;
}line[maxn*2];
struct sta
{
int x,y,value;
}star[maxn];
int lcount,dycount;
double dy[maxn*2];
map<double,int>map_m;
long long n,w,h;
bool cmp_double(double a,double b)
{
return a<b;
}
bool cmp_x(const li a,const li b)
{
if(a.x==b.x)
return !a.end;
else
return a.x<b.x;
}
void buildtree(int root,int L,int R)
{
tree[root].le=L;
tree[root].re=R;
tree[root].max_value=0;
tree[root].sing_value=0;
if(L!=R)
{
int mid = (L+R)/2;
buildtree(root*2+1,L,mid);
buildtree(root*2+2,mid+1,R);
}
}
void add_tree(int root,int L,int R,int c)
{
if(tree[root].le==L&&tree[root].re==R)
{
tree[root].sing_value+=c;
tree[root].max_value+=c;
return;
}
int mid = (tree[root].le+tree[root].re)/2;
if(R<=mid)
{
add_tree(root*2+1,L,R,c);
}
else if(L>=mid+1)
{
add_tree(root*2+2,L,R,c);
}
else
{
add_tree(root*2+1,L,mid,c);
add_tree(root*2+2,mid+1,R,c);
}
tree[root].max_value = max(tree[2*root+1].max_value,tree[2*root+2].max_value)+tree[root].sing_value;
}
int main()
{
//freopen("i.txt","r",stdin);
//freopen("o.txt","w",stdout);
int i,L1,R1;
long long ans;
double xx,yy;
while(scanf("%lld%lld%lld",&n,&w,&h)!=EOF)
{
xx=w/2.0;
yy=h/2.0;
map_m.clear();
lcount=0;
dycount=0;
for(i=1;i<=n;i++)
{
scanf("%d%d%d",&star[i].x,&star[i].y,&star[i].value);
line[lcount].x= star[i].x-xx;
line[lcount].y1=star[i].y-yy;
line[lcount].y2=star[i].y+yy;
line[lcount].value=star[i].value;
line[lcount].end=false;
lcount++;
line[lcount].x=star[i].x+xx;
line[lcount].y1=star[i].y-yy;
line[lcount].y2=star[i].y+yy;
line[lcount].value=star[i].value;
line[lcount].end=true;
lcount++;
dy[dycount++]=star[i].y-yy;
dy[dycount++]=star[i].y+yy;
}
sort(line,line+lcount,cmp_x);
sort(dy,dy+dycount,cmp_double);
dycount=unique(dy,dy+dycount)-dy;
buildtree(0,0,dycount-1);
for(i=0;i<dycount;i++)
{
map_m[dy[i]]=i;
}
ans=0;
for(i=0;i<lcount;i++)
{
L1=map_m[line[i].y1];
R1=map_m[line[i].y2];
if(!line[i].end)
{
add_tree(0,L1,R1,line[i].value);
ans=max(tree[0].max_value,ans);
}
else
{
add_tree(0,L1,R1,-1*line[i].value);
}
}
cout<<ans<<endl;
}
//system("pause");
return 0;
}
Stars in Your Window
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11073 | Accepted: 3033 |
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.
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.
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
POJ2482前面和题意无关，是一封非常优美的情书，觉得写得真是棒。
然后这个题目是不算边缘，一种做法就是建立线段树的时候像是之前的做法，叶子节点代表某一段空间了，另一种做法就是让窗户 减少一点点，在不影响圈星星的条件下，只减少一点点，这样就排除了边缘星星。
代码：
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#include <map>
#pragma warning(disable:4996)
using namespace std;
#define maxn 90002
struct no
{
int le,re;
long long max_value,sing_value;
}tree[maxn*5];
struct li
{
double x,y1,y2;
int value;
bool end;
}line[maxn*2];
struct sta
{
int x,y,value;
}star[maxn];
int lcount,dycount;
double dy[maxn*2];
map<double,int>map_m;
long long n,w,h;
bool cmp_double(double a,double b)
{
return a<b;
}
bool cmp_x(const li a,const li b)
{
if(a.x==b.x)
return !a.end;
else
return a.x<b.x;
}
void buildtree(int root,int L,int R)
{
tree[root].le=L;
tree[root].re=R;
tree[root].max_value=0;
tree[root].sing_value=0;
if(L!=R)
{
int mid = (L+R)/2;
buildtree(root*2+1,L,mid);
buildtree(root*2+2,mid+1,R);
}
}
void add_tree(int root,int L,int R,int c)
{
if(tree[root].le==L&&tree[root].re==R)
{
tree[root].sing_value+=c;
tree[root].max_value+=c;
return;
}
int mid = (tree[root].le+tree[root].re)/2;
if(R<=mid)
{
add_tree(root*2+1,L,R,c);
}
else if(L>=mid+1)
{
add_tree(root*2+2,L,R,c);
}
else
{
add_tree(root*2+1,L,mid,c);
add_tree(root*2+2,mid+1,R,c);
}
tree[root].max_value = max(tree[2*root+1].max_value,tree[2*root+2].max_value)+tree[root].sing_value;
}
int main()
{
//freopen("i.txt","r",stdin);
//freopen("o.txt","w",stdout);
int i,L1,R1;
long long ans;
double xx,yy;
while(scanf("%lld%lld%lld",&n,&w,&h)!=EOF)
{
xx=w/2.0-0.1;//！！！！！！！！！！！！！！！！！！！！这里有变化了
yy=h/2.0-0.1;//！！！！！！！！！！！！！！！！！！！！这里有变化了
map_m.clear();
lcount=0;
dycount=0;
for(i=1;i<=n;i++)
{
scanf("%d%d%d",&star[i].x,&star[i].y,&star[i].value);
line[lcount].x= star[i].x-xx;
line[lcount].y1=star[i].y-yy;
line[lcount].y2=star[i].y+yy;
line[lcount].value=star[i].value;
line[lcount].end=false;
lcount++;
line[lcount].x=star[i].x+xx;
line[lcount].y1=star[i].y-yy;
line[lcount].y2=star[i].y+yy;
line[lcount].value=star[i].value;
line[lcount].end=true;
lcount++;
dy[dycount++]=star[i].y-yy;
dy[dycount++]=star[i].y+yy;
}
sort(line,line+lcount,cmp_x);
sort(dy,dy+dycount,cmp_double);
dycount=unique(dy,dy+dycount)-dy;
buildtree(0,0,dycount-1);
for(i=0;i<dycount;i++)
{
map_m[dy[i]]=i;
}
ans=0;
for(i=0;i<lcount;i++)
{
L1=map_m[line[i].y1];
R1=map_m[line[i].y2];
if(!line[i].end)
{
add_tree(0,L1,R1,line[i].value);
ans=max(tree[0].max_value,ans);
}
else
{
add_tree(0,L1,R1,-1*line[i].value);
}
}
cout<<ans<<endl;
}
//system("pause");
return 0;
}
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