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 Input51 15 17 13 35 5Sample Output12110Hint
This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.
题意：
在坐标上有n个星星，如果某个星星坐标为(x, y), 它的左下位置为：(x0,y0)，x0<=x 且y0<=y。如果左下位置有a个星星，就表示这个星星属于level x
按照y递增，如果y相同则x递增的顺序给出n个星星，求出所有level水平的数量。
思路：求左下方的星星数量，而y是按照顺序排列的所有，找x0<=x 且y0<=y然后记录水平个数就好，如果暴力的话每个都要判断他前面的所有的星星是否符合条件，时间复杂度O（n*n）.需要找到一种数据结构来记录所有星星的x值，方便的求出所有值为0～x的星星总数量。可以按照x坐标建立一维树状数组.
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int MAXN=15010;
const int MAXX=32010;
int c[MAXX];//树状数组的c数组
int cnt[MAXN];//统计结果
int lowbit(int x)
{
return x&(-x);
}
void add(int i,int val)
{
while(i<=MAXX)
{
c[i]+=val;
i+=lowbit(i);
}
}
int sum(int i)
{
int s=0;
while(i>0)
{
s+=c[i];
i-=lowbit(i);
}
return s;
}
int main()
{
int n;
int x,y;
while(scanf("%d",&n)==1)
{
memset(c,0,sizeof(c));
memset(cnt,0,sizeof(cnt));
for(int i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
//加入x+1，是为了避免0，X是可能为0的
int temp=sum(x+1); //计算和（所处位置）
cnt[temp]++; //加1
add(x+1,1); //加入树状数组
}
for(int i=0;i<n;i++)
printf("%d\n",cnt[i]);
}
return 0;
}