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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

1
2
1
1
0
神啊原諒我吧我很菜！
看了樹狀數組還沒思路！
代碼ac后更新

posted on 2009-01-10 10:24 KNIGHT 閱讀(189) 評論(1) 編輯收藏引用

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# re: Stars[未登錄]

2009-01-12 10:01 | Knight

#include<stdio.h>
#define SIZE1 32000
#define SIZE2 15000
int c[SIZE1],a[SIZE1],out[SIZE2],n;
int lowbit(int k)
{
return k&(-k);
}
int sum(int k)
{
int ret=0;
while(k>0)
{
ret+=c[k];
k-=lowbit(k);
}
return ret;
}
void change(int pos,int delt)
{
while(pos<=SIZE1)
{
c[pos]+=delt;
pos+=lowbit(pos);
}
}
void init()
{
int i;
int x,y;
for(i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
x++;
out[sum(x-1)+a[x]]++;
change(x,1);
a[x]++;
}
}
int main()
{
int i;
scanf("%d",&n);
init();
for(i=0;i<n;i++)
printf("%d\n",out[i]);
}
pip來了，但是他說他也不會他是線段樹過的。。。。。
線段樹。。。。。代碼不是我的。。。網上的。。。。不過不錯。。。題目有個條件就是按y升序給出數據。。。。所以可以用樹狀數組，其實可以排序在用的。。。。。繼續物理。。。。關機。。。回復更多評論

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