描述
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.
输入
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.
输出
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.
样例输入
5
1 1
5 1
7 1
3 3
5 5
样例输出
1
2
1
1
0
提示
题目来源
做这题的时候一直超时。后来借了贞贞的代码看了，cjx更加混乱了，就两处地方改一改就不超时了。
实在搞不懂，求大神解答！
#include <stdio.h> #include <string.h> #define MAXN 32110 int n,m; int L[15010]; int C[MAXN]; int lowbit(int t){ return t&(t^(t-1)); } int sum(int end){ int sum(0); while(end>0){ sum+=C[end]; end-=lowbit(end); } return sum; } void add(int pos, int num){ while(pos<=MAXN){ C[pos]+=num; pos+=lowbit(pos); } } int main(int argc, char *argv[]) { int x,y; scanf("%d",&n); for(int i=1; i<=n; i++){ scanf("%d%d",&x,&y); x++; add(x,1); L[sum(x)]++; //----超时代码--- int level=sum(x); L[level]++; add(x,1); //--------------- } for(int i=1; i<=n; i++) printf("%d\n",L[i]); //----超时代码--- for(int i=0; i<n; i++) printf("%d\n",L[i]); //--------------- return 0; }
原文：http://www.cnblogs.com/chenjianxiang/p/3541147.html