hdoj 1541 Stars【线段树单点更新+最大值维护】

Stars

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6581 Accepted Submission(s):
2625


[align=left]Problem Description[/align]
Astronomers often examine star maps where stars are
represented by points on a plane and each star has Cartesian coordinates. Let
the level of a star be an amount of the stars that are not higher and not to the
right of the given star. Astronomers want to know the distribution of the levels
of the stars.



For example,
look at the map shown on the figure above. Level of the star number 5 is equal
to 3 (it's formed by three stars with a numbers 1, 2 and 4). And the levels of
the stars numbered by 2 and 4 are 1. At this map there are only one star of the
level 0, two stars of the level 1, one star of the level 2, and one star of the
level 3.

You are to write a program that will count the amounts of the
stars of each level on a given map.

[align=left]Input[/align]
The first line of the input file contains a number of
stars N (1<=N<=15000). The following N lines describe coordinates of stars
(two integers X and Y per line separated by a space, 0<=X,Y<=32000). There
can be only one star at one point of the plane. Stars are listed in ascending
order of Y coordinate. Stars with equal Y coordinates are listed in ascending
order of X coordinate.

[align=left]Output[/align]
The output should contain N lines, one number per line.
The first line contains amount of stars of the level 0, the second does amount
of stars of the level 1 and so on, the last line contains amount of stars of the
level N-1.

[align=left]Sample Input[/align]

5
1 1
5 1

7 1

3 3

5 5

[align=left]Sample Output[/align]

1

2

1

1

0

题意：一个星星的地图，每个星星都有对应的等级，星星等级的计算符合以下规律：一颗星星的 左下方（包括正左方和正下方）的星星个数就是这个星星的等级
题解：先对星星的纵坐标从小到大排列，如果纵坐标相同将横坐标从小到大排列，在进行数据输入时进行对此操作的处理，

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX 32010
using namespace std;
int sum[MAX<<2];
int pos[MAX];//记录各等级星星的个数
void pushup(int o)
{
sum[o]=sum[o<<1]+sum[o<<1|1];
}
void gettree(int o,int l,int r)
{
sum[o]=0;
if(l==r)
return ;
int mid=(l+r)>>1;
gettree(o<<1,l,mid);
gettree(o<<1|1,mid+1,r);
pushup(o);
}
void update(int o,int l,int r,int L)
{
if(l==r)
{
sum[o]+=1;
return ;
}
int mid=(l+r)>>1;
if(L<=mid)//因为建树的过程是按照星星的纵坐标从小到大建立的所以此处
//只需要考虑横坐标的情况，当横坐标小于mid对左子树进行操作
//否则对右子树进行操作。
update(o<<1,l,mid,L);
else
update(o<<1|1,mid+1,r,L);
pushup(o);
}
int find(int o,int l,int r,int L,int R)
{
if(L<=l&&R>=r)
return sum[o];
int ans=0;
int mid=(l+r)>>1;
if(L<=mid)
ans+=find(o<<1,l,mid,L,R);
if(R>mid)
ans+=find(o<<1|1,mid+1,r,L,R);
return ans;
}
int main()
{
int n,m,j,i;
int x,y;
int level;
while(scanf("%d",&n)!=EOF)
{
memset(pos,0,sizeof(pos));
gettree(1,1,MAX);
for(i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
level=find(1,1,MAX,1,x+1);
pos[level]++;
update(1,1,MAX,x+1);
}
for(i=0;i<n;i++)
printf("%d\n",pos[i]);
}
return 0;
}