【题解】POJ2352：Stars

原题传送门
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

【题目大意】
平面直角坐标系上有n个点，每个点有一个等级，第i个点的等级为坐标系上x轴、y轴坐标均小于等于自己的点（不包括自己）的个数
输出等级为0~n-1分别有几个点

【题解】
明确等级的意思，对于某个点，过它向x、y轴作垂线，和坐标轴围成的一个矩形包含的点的个数（矩形边上算进，自己不算）即为等级

二维前缀和？
显然不行，这个范围在32000之内，肯定RE
先离散化一下？
也不行，n也有15000，运气差一点也RE了
那怎么办？

发现输入数据是按y坐标从小到大输入的
那么y的单调性已经具备了，只用考虑x坐标就行了！
所以明确算法树状数组！！
按照x坐标累计树状数组，当然要注意到：0<=X,Y
但树状数组不能有零！！
为什么？想象一下，当循环的执行条件为x>=0，所以x=0的时候循环也在执行的，但x-=lowbit(x)这个操作就hh了，因为0-0&-0=0，所以会陷入死循环
那么我们再把x坐标全部+1就行了

这道题线段树也可以，但效率跟树状数组差不多。
树状数组在这里就有绝对优势了——码量小！！

Code:

#include <cstdio>
#include <algorithm>
#define maxn 32010
#define ll long long
#define res register int
using namespace std;
int tree[maxn],n,print[maxn];

inline int read(){
    int s = 0,w = 1;
    char c = getchar();
    while (c < '0' || c > '9'){if (c == '-') w = -1; c = getchar();}
    while (c >= '0' && c <= '9') s = s * 10 + c - '0',c = getchar();
    return s * w;
}
inline void change(int x){for (;x <= maxn;x += (x & -x)) ++ tree[x];}
inline int getsum(int x){int sum = 0; for (;x > 0;x -= (x & -x)) sum += tree[x]; return sum;}
int main(){
    n = read();
    for (res i = 1;i <= n;++ i){
        int x = read(), y = read();
        int sum = getsum(x + 1);
        ++ print[sum];
        change(x + 1);
    }
    for (res i = 0;i < n;++ i) printf("%d\n",print[i]);
    return 0;
}