Stars（BIT树状数组）

Stars

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6993 Accepted Submission(s):
2754


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

5
1 1
5 1

7 1

3 3

5 5

[align=left]Sample Output[/align]

1
2
1
1
0
给你一堆星星的坐标，一个星星的level等于它左下边的星星个数，不包括自己本身，可同x,同y;
而且题目给的星星的顺序是按照y,x坐标来排序的，BIT搞。

/*******************************

Date    : 2015-12-09 23:16:34
Author  : WQJ (1225234825@qq.com)
Link    : http://www.cnblogs.com/a1225234/ Name     :

********************************/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
int bit[32000+5];
int num[32000+5];
int k=32000+5;
int lowbit(int i)
{
return i&-i;
}
void add(int i,int a)
{
while(i<=32005)
{
bit[i]+=a;
i=i+lowbit(i);
}
}
int sum(int i)
{
int s=0;
while(i>0)
{
s+=bit[i];
i=i-lowbit(i);
}
return s;
}
int main()
{
freopen("in.txt","r",stdin);
int i,j;
int n,x,y;
while(scanf("%d",&n)!=EOF)
{
memset(num,0,sizeof(num));
memset(bit,0,sizeof(bit));
for(i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
num[sum(x+1)]++;
add(x+1,1);
}
for(i=0;i<n;i++)
printf("%d\n",num[i]);
}
return 0;
}

据说暴力也能过：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[33000],i,j,k,l,m,n,x[33000],y[33000];
int main()
{
while(scanf("%d",&k)!=EOF)
{
memset(a,0,sizeof(a));
for(j=0;j<k;j++)
{
scanf("%d%d",&x[j],&y[j]);
int cnt=0;
for(l=0;l<j;l++)
if(x[l]<=x[j])
cnt++;
a[cnt]++;
}
for(i=0;i<k;i++)
printf("%d\n",a[i]);

}
return 0;
}