PKU 2352 Stars

题目：http://poj.org/problem?id=2352

题意：平面上有n个星星，每个星星的等级是比该星星的X和Y坐标都小的星星的个数,给出每个星星的坐标（按y轴升序），输出等级0~n-1每个等级的星星个数。

思路：线段树，因为y轴升序给出，我们只考虑x轴就可以， 每次查询只需要查询0～x区间的个数即可。

#include <stdio.h>
#include <string.h>
const int maxn=32000;
struct SegMentTree
{
	int l, r, sum;
}st[maxn<<2];
int lvl[maxn+10];
int n, x, y;
void BuildTree(int left, int right, int v)
{
	st[v].l=left, st[v].r=right;
	if (left==right)
	{
		st[v].sum=0;
		return;
	}
	int lc=v<<1;
	int rc=lc+1;
	int mid=(left+right)>>1;
	BuildTree(left, mid, lc);
	BuildTree(mid+1, right, rc);
	st[v].sum=st[lc].sum+st[rc].sum;
}
void Insert(int left, int right, int v)
{
	if (st[v].l==left&&st[v].r==right)
	{
		st[v].sum++;
		return;
	}
	int lc=v<<1;
	int rc=lc+1;
	int mid=(st[v].l+st[v].r)>>1;
	if (mid>=right) Insert(left, right, lc);	
	else if (left>mid) Insert(left, right, rc);
	else
	{
		Insert(left, mid, lc);
		Insert(mid+1, right, rc);
	}
	st[v].sum=st[lc].sum+st[rc].sum;
}
int Query(int left, int right, int v)
{
	if (st[v].l==left&&st[v].r==right) return st[v].sum;
	int lc=v<<1;
	int rc=lc+1;
	int mid=(st[v].l+st[v].r)>>1;
	if (mid>=right) return Query(left, right, lc);
	else if (left>mid) return Query(left, right, rc);
	else return Query(left, mid, lc)+Query(mid+1, right, rc);
}
int main()
{
	//freopen("in.txt", "r", stdin);
	while (scanf("%d", &n)==1)
	{
		memset(lvl, 0, sizeof(lvl));
		BuildTree(0, maxn, 1);
		for (int i=0; i<n; i++)
		{
			scanf("%d %d", &x, &y);
			lvl[Query(0, x, 1)]++;
			Insert(x, x, 1);
		}
		for (int i=0; i<n; i++) printf("%d\n", lvl[i]);
	}
}