POJ 2352-Stars
Stars
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 36453 Accepted: 15852
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
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
星星的level为其左下星星的数量，要求输出level从0~n-1的星星的个数。 因为本题的数据已经按先Y后X从小到大排好序了，所以后面的星星一定不可能出现在前面星星的左下部分，所以当来了一个星星(x,y)之后，只需求出之前出现过的星星中x'<=x的星星的个数。 所以可以只考虑星星的x坐标。假设有一个数组A[]，A[x]表示横坐标为x的星星的个数，则可以画出这样一张图： [caption id="" align="alignnone" width="499"]
树状数组[1][/caption]当来了一个横坐标为4的星星之后，其level正是A[1]+…+A[4]=C[4]，在树状数组中正好是其左下方的和。
关于树状数组可以参考[1]的文章。
所以横坐标为4的星星的level就是sum(4)。当又来了一个横坐标为4的星星之后，需要修改A[4]，使A[4]++，其实就是A[4]+1，可以使用add(4,1)操作。
因为本题中x可能等于0，但是树状数组中下标从1开始，所以需要整体将x+1。本题是树状数组的一个简单应用，完整代码如下：
[cpp]
#include<iostream>
#include<cstdio>
using namespace std;
const int kMaxN = 32005;//NOT 15005;
int n;
int level[kMaxN];
int c[kMaxN];
int lowbit(int x)
{
return x&(-x);
}
int sum(int i)
{
int rs = 0;
while (i > 0)
{
rs += c[i];
i -= lowbit(i);
}
return rs;
}
void add(int i, int val)
{
while (i <= kMaxN)
{
c[i]+=val;
i += lowbit(i);
}
}
int main()
{
scanf("%d", &n);
int x, y;
for (int i = 0; i < n; i++)
{
scanf("%d %d", &x, &y);
x++;
level[sum(x)]++;
add(x, 1);
}
for (int i = 0; i < n; i++)
printf("%dn", level[i]);
return 0;
}
[/cpp]
本代码提交AC，用时188MS，内存344K。
参考：
[1]. http://www.cnblogs.com/Creator/archive/2011/09/10/2173217.html
]]>