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.
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
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题意是说，有n颗星星，排列在笛卡尔坐标中，输入时按y坐标递增，y坐标相等时，按x坐标递增。
求最后每颗星星左下方的星星数
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1 #include <iostream> 2 #include <string> 3 #include <cstring> 4 #include <queue> 5 #include <vector> 6 #include <cstdio> 7 #define lowbit(x) ((x)&(-(x))) 8 using namespace std; 9 const int maxn=15007; 10 const int maxx=32002; 11 int bit[maxx],level[maxn]; 12 int n; 13 void add(int pos,int delt) 14 { 15 for(int i=pos;i<=maxx;i+=lowbit(i)) 16 bit[i]+=delt; 17 } 18 19 int sum(int pos) 20 { 21 int re=0; 22 for(int i=pos;i>0;i-=lowbit(i)) 23 { 24 re+=bit[i]; 25 } 26 return re; 27 } 28 29 int main() 30 { 31 while(scanf("%d",&n)==1) 32 { 33 memset(bit,0,sizeof(bit)); 34 memset(level,0,sizeof(level)); 35 for(int i=1;i<=n;i++) 36 { 37 int x,y; 38 scanf("%d%d",&x,&y); 39 level[sum(x+1)]++; 40 add(x+1,1); 41 } 42 for(int i=0;i<n;i++) 43 printf("%d\n",level[i]); 44 } 45 return 0; 46 }