HDU 1541 树状数组

Stars

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10583    Accepted Submission(s): 4230

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

51 15 17 13 35 5

 

Sample Output

12110

 

题意：求每个星星的等级，其实就是求每个星星左下角有多少个星星，然后输出每个等级有多少颗星星

思路：题目有个特殊的地方，就是星星的输入是 y相同的时候x从小到大输入，然后y是从小到大的。

所以对于当前的星星的输入，它一定是目前为止同高度下x最大的，也就是说，它左下角的星星数目已经是确定了。

所以每输入一个星星就可以去计算它的等级了。用树状数组，以横坐标为编号（纵坐标也行）维护树状数组就可以了

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define mem(a,x) memset(a,x,sizeof(a))#define lowbit(x) (x & (-x))#define maxn 32005int tree[maxn],ans[maxn];void update(int pos,int val){while(pos <= maxn){tree[pos] += val;pos += lowbit(pos);}}int getSum(int x){int sum = 0;while(x > 0){sum += tree[x];x -= lowbit(x);}return sum;}int main(){int n,x,y;while(scanf("%d",&n) != EOF){mem(tree,0);mem(ans,0);for(int i = 1;i <= n;i++){scanf("%d %d",&x,&y);ans[getSum(x + 1)]++;update(x + 1,1);}for(int i = 0;i < n;i++){printf("%d\n",ans[i]);}}return 0;}