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.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.Source
Ural Collegiate Programming Contest 1999
level就是指當前星星的左下方的星星個數。因為y坐標已經排序,則對於每個星星只需要求x坐標前的星星個數了,典型的樹狀數組求解。因為x可以等於0,所以我把每個x坐標都+1處理。
AC代碼:
#include#include #include using namespace std; int n; int c[320005],d[15005]; int lowbit(int x){ return x&-x; } int sum(int x){ int ret=0; while(x>=1){ ret+=c[x]; x-=lowbit(x); } return ret; } void add(int x){ while(x<=320001){ c[x]++; x+=lowbit(x); } } int main(){ while(scanf("%d",&n)!=EOF){ memset(c,0,sizeof(c)); memset(d,0,sizeof(d)); for(int i=1;i<=n;i++){ int x,y; scanf("%d%d",&x,&y); d[sum(x+1)]++; add(x+1); } for(int i=0;i