poj 2352 && hdu 1541 Stars 线段树
2016-10-25 12:04
597 查看
题意:
按序在图上插入点,求每次插入点左下方有几个点,点的个数即为此点的等级。输出每个等级点的个数。
思路:
由于插入点的顺序是先下后上的
故可以x轴为准,先查询点的左边有几个点,再插入点,不用考虑y轴。
代码:
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
#define ls rt<<1,l,mid
#define rs rt<<1|1,mid+1,r
const int MAXN=40000;
int st,en;
int tree[MAXN*3];
void push_up(int rt){
tree[rt]=tree[rt<<1]+tree[rt<<1|1];
return ;
}
void update(int rt,int l,int r){
if(l==r){
tree[rt]=1;
return ;
}
int mid=(l+r)>>1;
if(en<=mid)
update(ls);
if(mid<en)
update(rs);
push_up(rt);
}
int query(int rt,int l,int r){
if(st<=l&&r<=en)
return tree[rt];
int ans=0;
int mid=(l+r)>>1;
if(st<=mid)
ans+=query(ls);
if(mid<en)
ans+=query(rs);
return ans;
}
typedef struct Node{
int x;
int y;
int id;
bool operator <(const Node& a)const{
if(a.x==x)
return a.y>y;
return a.x>x;
}
}Node;
int main()
{
int n;
while(cin>>n){
memset(tree,0,sizeof(tree));
Node a[MAXN];
int ans[MAXN]={0};
//int m=0;
for(int i=0;i<n;i++){
scanf("%d %d",&a[i].x,&a[i].y);
a[i].id=i+1;
//m=max(m,a[i]);
}
sort(a,a+n);
for(int i=0;i<n;i++){
st=1;en=a[i].id;
ans[query(1,1,n)]++;
update(1,1,n);
}
for(int i=0;i<n;i++)
printf("%d\n",ans[i]);
}
return 0;
}
Description
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.
Input
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.
Output
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.
Sample Input
Sample Output
Hint
This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.
按序在图上插入点,求每次插入点左下方有几个点,点的个数即为此点的等级。输出每个等级点的个数。
思路:
由于插入点的顺序是先下后上的
故可以x轴为准,先查询点的左边有几个点,再插入点,不用考虑y轴。
代码:
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
#define ls rt<<1,l,mid
#define rs rt<<1|1,mid+1,r
const int MAXN=40000;
int st,en;
int tree[MAXN*3];
void push_up(int rt){
tree[rt]=tree[rt<<1]+tree[rt<<1|1];
return ;
}
void update(int rt,int l,int r){
if(l==r){
tree[rt]=1;
return ;
}
int mid=(l+r)>>1;
if(en<=mid)
update(ls);
if(mid<en)
update(rs);
push_up(rt);
}
int query(int rt,int l,int r){
if(st<=l&&r<=en)
return tree[rt];
int ans=0;
int mid=(l+r)>>1;
if(st<=mid)
ans+=query(ls);
if(mid<en)
ans+=query(rs);
return ans;
}
typedef struct Node{
int x;
int y;
int id;
bool operator <(const Node& a)const{
if(a.x==x)
return a.y>y;
return a.x>x;
}
}Node;
int main()
{
int n;
while(cin>>n){
memset(tree,0,sizeof(tree));
Node a[MAXN];
int ans[MAXN]={0};
//int m=0;
for(int i=0;i<n;i++){
scanf("%d %d",&a[i].x,&a[i].y);
a[i].id=i+1;
//m=max(m,a[i]);
}
sort(a,a+n);
for(int i=0;i<n;i++){
st=1;en=a[i].id;
ans[query(1,1,n)]++;
update(1,1,n);
}
for(int i=0;i<n;i++)
printf("%d\n",ans[i]);
}
return 0;
}
Description
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.
Input
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.
Output
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.
Sample Input
5 1 1 5 1 7 1 3 3 5 5
Sample Output
1 2 1 1 0
Hint
This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.
相关文章推荐
- POJ 2352 && HDU 1541 Stars(BIT)
- poj 2352 && hdu 1541 Stars (树状数组水题)
- poj-2352 && HDU-1541 --Stars(树状数组)
- poj 2352 && hdu 1541 Stars (树状数组)
- poj 2352 && hdu 1541 Stars(树状数组)
- POJ 2352 && HDU 1541 Stars (树状数组)
- poj 2352 && hdu 1541 Stars (树状数组)
- HDU 1541 & POJ 2352 Stars (树状数组)
- POJ 2352 && HDU 1541 Stars (树状数组)
- PKU 2481 Cows & PKU 2352 Stars & HDU 1541 Stars
- HDU 1541+poj 2352 stars
- POJ 2352 Stars(HDU 1541 Stars)
- hdu 1541/poj 2352:Stars(树状数组,经典题)
- HDU 1541 Stars || POJ 2352 stars || NYOJ 117 求逆序数
- POJ2352 Stars(线段树 & 树状数组)
- POJ 2352_Stars && POJ-2481 Cows (线段树单点更新+树状数组)
- 树状数组 POJ 2352 HDU 1541 Stars
- POJ 2352 Stars Treap & 线段树
- POJ 2777 && ZOJ 1610 &&HDU 1698 --线段树--区间更新
- POJ 2299 && HDU 3743 离散化+线段树