Stars
| Time Limit: 1000MS | | Memory Limit: 65536K |
| Total Submissions: 14814 | | Accepted: 6404 |
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
與樹狀數(shù)組的解法一樣, 查詢更新就可以了.
樹狀數(shù)組 解法傳送門 : http://www.cnblogs.com/MiYu/archive/2010/08/26/1808963.html
代碼如下 :
/*
Mail to : miyubai@gamil.com
Link : http://www.cnblogs.com/MiYu || http://www.shnenglu.com/MiYu
Author By : MiYu
Test : 1
Complier : g++ mingw32-3.4.2
Program : POJ_2352
Doc Name : Stars
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
struct P{
int x, y;
}p[15010];
int seg[100000];
int NN = 32768, res[15010];
int get ( int &n ) {
int i = 1;
while ( i < n ) i << 1;
return i;
}
void modify ( int &pos ) {
pos += NN;
seg[pos] ++;
while ( pos > 1 ) {
if ( ~pos & 1 ) seg[pos>>1] ++;
pos >>= 1;
}
}
int query ( int pos ) {
pos += NN;
int sum = seg[pos];
while ( pos > 1 ) {
if ( pos & 1 ) {
sum += seg[pos>>1];
}
pos >>= 1;
}
return sum;
}
int main ()
{
int N;
while ( scanf ( "%d", &N ) == 1 ) {
int pos;
memset ( res, 0, sizeof ( res ) );
memset ( seg, 0, sizeof ( seg ) );
for ( int i = 0; i < N; ++ i ) {
scanf ( "%d%*d", &pos );
res[ query ( pos ) ] ++;
modify ( pos );
}
for ( int i = 0; i < N; ++ i ) {
printf ( "%d\n", res[i] );
}
}
return 0;
}