【线段树】Stars
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1028. Stars
Time Limit: 0.25 second
Memory Limit: 16 MB
Memory Limit: 16 MB
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 contains a number of starsN (1 ≤ N ≤ 15000). The following N lines describe coordinates of stars (two integersX 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 ofY coordinate. Stars with equal Y coordinates are listed in ascending order ofX 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
51 15 17 13 35 5
12110
应该是很久以前做的题，但是现在才做。SS一定要把它讲成是线段树和树状数组的过渡品，我不置可否。
原来看题的时候，它是笛卡尔坐标系上的，不知如何处理，但是今天再看题，就想到了当任意一维坐标有序之后，就只需要考虑另一维勒。
于是我排序之后就不知如何办了。。结果后来再审题发现本来数据就是有序的。。。。
以当前读入的点为原点建系，所有已有点在三四象限。而第三象限是满足要求的，因此只和横坐标有关了。
解释一下程序：
1、没有建树，也许建树要快一点，因为这里范围有点大，1~32000,。
2、这里把查找和插入合并了，tree维护的是1~r这个区间点的数量
总之可以由这道题发散思维，这道题的线段树应用得应该是很灵活的，不拘谨。
其实当想到排序之后，仅仅和另一维有关，要统计某点左边的点数量，应该很容易想到树状数组
#include<cstdio>#include<cstdlib>using namespace std;long k;long cnt[15010];long tree[128010];void insert(long l,long r,long i,long x){long mid = (l+r)>>1;if (l == r){k += tree[i];tree[i] ++;return;}if (x <= mid){insert(l,mid,i<<1,x);tree[i] ++;}else {insert(mid+1,r,(i<<1)+1,x);k += tree[i<<1];tree[i] ++;}}int main(){long n;scanf("%ld",&n);for (long i=1;i<n+1;i++){long x;k = 0;long tmp;scanf("%ld%ld",&x,&tmp);insert(1,32000,1,x);cnt[k] ++;}for (long i=0;i<n;i++){printf("%ld\n",cnt[i]);}return 0;}
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