POJ2352_Stars(线段树/单点更新)
2014-08-09 19:15
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解题报告
题意:
坐标系中,求每颗星星的左下角有多少星星。
思路:
把横坐标看成区间,已知输入是先对y排序再对x排序,每次加一个点先查询该点x坐标的左端有多少点,再更新点。
Stars
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.
![](http://poj.org/images/2352_1.jpg)
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.
题意:
坐标系中,求每颗星星的左下角有多少星星。
思路:
把横坐标看成区间,已知输入是先对y排序再对x排序,每次加一个点先查询该点x坐标的左端有多少点,再更新点。
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int sum[200000]; struct node { int x,y; } p[20000]; void push_up(int root) { sum[root]=sum[root*2]+sum[root*2+1]; } void update(int root,int l,int r,int p,int v) { int mid=(l+r)/2; if(l==r) { sum[root]++; return; } if(p<=mid)update(root*2,l,mid,p,v); else update(root*2+1,mid+1,r,p,v); push_up(root); } int q_sum(int root,int l,int r,int ql,int qr) { if(ql>r||qr<l)return 0; if(ql<=l&&r<=qr)return sum[root]; int mid=(l+r)/2; return q_sum(root*2,l,mid,ql,qr)+q_sum(root*2+1,mid+1,r,ql,qr); } int main() { int n,i,j,m=32000; int _hash[20000]; scanf("%d",&n); memset(_hash,0,sizeof(_hash)); for(i=0; i<n; i++) { scanf("%d%d",&p[i].x,&p[i].y); _hash[q_sum(1,0,m,0,p[i].x)]++; update(1,0,m,p[i].x,1); } for(i=0; i<n; i++) printf("%d\n",_hash[i]); return 0; }
Stars
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 32329 | Accepted: 14119 |
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.
![](http://poj.org/images/2352_1.jpg)
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.
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