[ACM_資料結構] POJ2352 [樹狀數組稍微變形]

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

      

1 #include<iostream>
 2 #include<cstdio>
 3 #include<memory.h>
 4 using namespace std;
 5 #define maxn 32100//是葉節點能達到的最大值多一點
 6 int C[maxn];//從1開始編号
 7 //--------------------------
 8 int lowbit(int x){
 9     return x&-x;
10 }
11 int sum(int x){
12     int ret=0;
13     while(x>0){
14         ret+=C[x];
15         x-=lowbit(x);
16     }
17     return ret;
18 }
19 void add(int x,int d){
20     while(x<=maxn){
21         C[x]+=d;
22         x+=lowbit(x);
23     }
24 }
25 //--------------------------
26 /*
27 給定n個點，問這一點左下角的區域的點的個數，不包括這一點
28 因為輸入是按照y從小到達，然後是按x從小到達，且x不同
29 是以先出現的肯定比後出現的y小即在後出現的點之下，
30 是以隻要把x作為樹狀數組下标就可以啦。然後用一個out[num]維護
31 個數為num的點的數量
32 */
33 int out[maxn];
34 int main(){
35     int n;
36     while(scanf("%d",&n)!=EOF){
37         int x,y;
38         int i,j;
39         memset(out,0,sizeof(out));
40         memset(C,0,sizeof(C));
41         for(i=1;i<=n;i++){
42             scanf("%d%d",&x,&y);
43             out[sum(x+1)]++;
44             add(x+1,1);
45         }
46         for(i=0;i<n;i++){
47             printf("%d\n",out[i]);
48         }
49     }
50     return 0;
51 }      

http://www.cnblogs.com/zjutlitao/p/3701986.html