POJ 2352 数星星

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 53816		Accepted: 23159

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

Sample Output

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

****这是一道树状数组的题，根据题目可知，星星位置的顺序已经排好序了，所以对于本题来说y坐标的位置没有用，只需要一个一个星星判断他们的x坐标位置即可，由于不确定级数的范围大小，所以把级数直接设为最大可能行32001。由于树状数组的角标不能是0，所以把每个坐标的x值都加一，当然也可以加别的数字，但会增大数据范围。

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<algorithm>
 4 #include<cstring>
 5 using namespace std;
 6 int n,m,i,j,k,u,v,ans[15005] ={0},tree[32001] = {0};
 7 struct node
 8 {
 9     int x;
10     int y;
11 }a[15005];
12 int lowbit(int x)
13 {
14     return x & (-x);
15 }
16 void add(int l,int x)
17 {
18     while(l <= 32001)
19     {
20         tree[l] += x;
21         l += lowbit(l);
22     }
23 }
24 int sum(int p)
25 {
26     int res = 0;
27     while(p > 0)
28     {
29         res += tree[p];
30         p -= lowbit(p);
31     }
32     return res;
33 }
34 int main()
35 {
36     scanf("%d",&n);
37     for(i = 1;i <= n;i++)
38     {
39         scanf("%d %d",&a[i].x,&a[i].y);
40     }
41     for(i = 1;i <= n;i++)
42     {
43         u = a[i].x + 1;
44         v = sum(u);
45         ans[v] += 1;
46         add(u,1);
47     }
48     for(i = 0;i < n;i++)
49     {
50         printf("%d",ans[i]);
51         if(i != n)
52             printf("\n");
53     }
54     return 0;
55 }

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