Stars - POJ 2352 树状数组

最新推荐文章于 2021-05-03 20:08:45 发布

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本文介绍了一道关于星图统计的问题，通过使用树状数组来高效地统计每个星星的级别分布。输入包括星星数量及坐标，输出为不同级别的星星数量。

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30686		Accepted: 13405

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

题意：给你一组数，代表星星的位置，这个星星左下方有多少个星星，那么这个星星的value就是多少，问value从0到n-1各有多少个。

思路：简单的树状数组。数据是按照顺序给的，所以不用排序，另外此题不小心就会TLE原因是给的数有0，把么lowbit（0）=0，它就会一直循环下去，每次让x++即可。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;

int s[32010];
int value[15010];
int lowbit(int x)
{ return x&(-x);
}
void update(int x)
{ int i;
  for(i=x;i<=32001;i+=lowbit(i))
    s[i]++;
}
int sum(int x)
{ int res=0,i;
  for(i=x;i>0;i-=lowbit(i))
     res+=s[i];
  return res;
}
int main()
{ int T,x,y,i;
  memset(s,0,sizeof(s));
  scanf("%d",&T);
  for(i=T;i>0;i--)
  { scanf("%d%d",&x,&y);
    x++;
    value[sum(x)]++;
    update(x);
  }
  for(i=0;i<T;i++)
  { printf("%d\n",value[i]);
  }
}