POJ-2352Stars(BIT|归并排序求逆序对)

Stars
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 48588 Accepted: 20999
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
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

题意：

题目大意：给n个，每个都一个坐标（x,y)，问对于每个* ，其左下方共有多少个* （包含左下方同行，同列的* ）

对y坐标从小到大排序，如果y相同，就暗x坐标从小到大排序。相当于按x轴建立一维树状数组，然后求相当于它前面比它小的个数即可。
就转化为逆序对。。

#include
#include
#include
#include
using namespace std;
typedef long long LL;
const int maxn=1e5+5,N=32000+5;

int T[N],ans[maxn],x,y,n; 

inline int get(int x)
{
  int res=0;
  for(;x;x-=x&-x)res+=T[x];
  return res;
}

inline void add(int x)
{
  for(;xint main()
{
  while(~scanf("%d",&n))
  {
    memset(ans,0,sizeof(ans));
    memset(T,0,sizeof(T));
    for(int i=1;i<=n;++i)
    {
      scanf("%d%d",&x,&y);
      int res=get(x+1);
      ans[res]++;
      add(x+1);
    }
    for(int i=0;iprintf("%d\n",ans[i]);
  }
  return 0;
}