StarsTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5378Accepted Submission(s): 2125
Problem 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
51 15 17 13 35 5
Sample Output
12110
记得以前做这题对的时候，不会做，就留着没做。
如今，，我已经A了这题。顿时感觉以前的自己好菜。。
输入的部分：Stars are listed in ascending order of Y coordinate. Stars with equal Y coordinates are listed in ascending order of X coordinate.
人家都已经帮排好序了。。我以前还不会做。。。
代码：
#include <cstdio>#include <cstring>#define MAX 20000#define LEN 32500using namespace std ;struct Point{int x, y ;}p[MAX] ;int t[LEN] , level[MAX] ;int lowbit(int n){return n&(-n) ;}void update(int pos){if(pos == 0)return ;while(pos<LEN){t[pos]++ ;pos += lowbit(pos) ;}}int sum(int pos){int s = 0 ;while(pos>0){s += t[pos] ;pos -= lowbit(pos) ;}return s ;}int main(){int n ;while(~scanf("%d",&n)){memset(level,0,(n+10)*sizeof(int)) ;memset(t,0,sizeof(t)) ;for(int i = 0 ; i < n ; ++i){scanf("%d%d",&p[i].x , &p[i].y) ;p[i].x++ ;}for(int i = 0 ; i < n ; ++i){level[sum(p[i].x)]++ ;update(p[i].x) ;}for(int i = 0; i < n ; ++i){printf("%d\n",level[i]) ;}}return 0 ;}与君共勉
以一种进取的和明智的方式同它们奋斗。
