树状数组的入门题，做完之后发现还是没有变形，依然套模板。
这题还要对数据处理一下。树状数组的值是从一开始时的，因为0 要用来判断结束所以+1；因为不算自己所以先求和。
Stars
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7266 Accepted Submission(s): 2847
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.
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
Source
Recommend
LL
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 const int size = 32010; 7 const int N = 15010; 8 int x; 9 int c[size]; 10 int ligh[N]; 11 int lowbit(int x){ 12 return x&(-x); 13 } 14 void add(int k){ 15 while(k<=size){ 16 c[k]++; 17 k+=lowbit(k); 18 } 19 } 20 int sum(int k){ 21 int sum=0; 22 while(k>0){ 23 sum+=c[k]; 24 k-=lowbit(k); 25 } 26 return sum; 27 } 28 int main(){ 29 int n,y; 30 while(scanf("%d",&n)!=EOF){ 31 for(int i=0;i<n;i++) ligh[i]=0; 32 memset(c,0,sizeof(c)); 33 for (int i=0;i<n;i++){ 34 scanf("%d%d",&x,&y); 35 ligh[sum(x+1)]++; 36 add(x+1); 37 } 38 for(int i=0;i<n;i++){ 39 printf("%d ",ligh[i]); 40 } 41 } 42 return 0; 43 }