Stars

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8460    Accepted Submission(s): 3383 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 5 1 1 5 1 7 1 3 3 5 5   Sample Output 1 2 1 1 0  

这题技巧性太强x,y的输入顺序，还要注意细节，只要能联想到树状数组就可以把x的坐标当成区间端点，利用求和特性求出

#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int N = 32110; int c[N], total[N]; int lowbit(int k) {     return k&-k; } void add(int x) {     while(x<=N)     {         c[x]+=1;         x+=lowbit(x);     }     return ; } int sum(int x) {     int ans=0;     while(x>0)     {         ans+=c[x];         x-=lowbit(x);     }     return ans; } int main() {     int n;     while(scanf("%d", &n)!=EOF)     {         memset(c,0,sizeof(c));         memset(total,0,sizeof(total));         int x, y;         for(int i=0; i<n; i++)         {             scanf("%d %d", &x, &y);             add(x+1);

由于坐标x可能为0，因此输入坐标要+1，不然会超时0&(-0)=0;             total[sum(x+1)-1]++;         }         for(int i=0; i<n; i++)         {             printf("%d\n",total[i]);         }     }     return 0; }