Stars
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
Hint
This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.
题目的大体意思是：输出在每颗星星左下侧的星星的数量。
最直接的方法就是开一个二维数组，大范围搜索，这样的朴素算法不但会超时，在最坏的情况下数组会开到很大（具体多大懒得算了）
能不能把数据排序，减少计算次数呢？如果排序，按照横坐标还是纵坐标呢？
通过尝试，我们选择对纵坐标排序，用一维数组在每颗星的位置标为1，最终第i颗星就有
个。
这种算法的时间复杂度O(N2)，很遗憾还是会超时。
对于多次的数据更新，我们可以用非常优秀的一种数据结构——树状数组，时间复杂度只有O(logN)。
例如这张图，顺序是1->2->4->3。
代码如下：
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 int n,s[40000]={},answer[40000],ans,maxx=0; 5 int lowbit(int x) {return x & -x;} 6 7 struct node{ 8 int x,y; 9 }a[20000]; 10 11 int cmp(node a,node b){ 12 return a.y<b.y; 13 } 14 15 void input(){ 16 for(int i=1;i<=n;i++){ 17 scanf("%d%d",&a[i].x,&a[i].y); 18 a[i].x+=1; //防止0的出现 19 20 if(maxx<a[i].x) maxx=a[i].x; 21 } 22 } 23 24 int main(){ 25 // freopen("poj2352.in","r",stdin); 26 scanf("%d",&n); 27 input(); 28 sort(a,a+n,cmp); 29 for(int i=1;i<=n;i++){ 30 ans=0; 31 for(int j=a[i].x;j;j-=lowbit(j)){ 32 ans+=s[j]; 33 } 34 for(int j=a[i].x;j<=maxx;j+=lowbit(j)){ 35 s[j]++; 36 } 37 answer[ans]++; 38 } 39 40 for(int i=0;i<n;i++) printf("%d\n",answer[i]); 41 42 return 0; 43 }