Stars 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 st…

题目链接 题意:在一个二维平面上有n个星星,每个星星的等级为x,x为该星星左方和下方所包含的星星的数量(包含正左和正下的),输出每个等级各有多少星星,星星坐标按照y序递增给出,y值相同按照x递增给出. 题解:因为已经排好序了,我们每次更新加查询就可以了,因为后加入的一定是下方或者同行的,查询一下是不是左面的就行了.可以用线段树或者树状数组做,注意建树是N不是n,区间更新问题好像可以用什么lazy标记,这道题主要考察的还是思路.注意本题是按照值来进行查询的,建树的时候要用N建树. #include…

转载自 http://www.cnblogs.com/fenshen371/archive/2013/07/25/3214927.html 题意:已知n个星星的坐标.每个星星都有一个等级,数值等于坐标系内纵坐标和横坐标皆不大于它的星星的个数.星星的坐标按照纵坐标从小到大的顺序给出,纵坐标相同时则按照横坐标从小到大输出. (0 <= x, y <= 32000) 要求输出等级0到n-1之间各等级的星星个数.详细信息点击进入传送门. 思路:这个题目数据的输出顺序给了很大的便利.由于是按照纵坐标从小…

Stars Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 30496   Accepted: 13316 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 st…

Stars Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4052    Accepted Submission(s): 1592 Problem Description Astronomers often examine star maps where stars are represented by points on a plan…

Language:Default Stars Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 52268 Accepted: 22486 Description Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the…

Stars Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 54352   Accepted: 23386 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 st…

<题目链接> 题目大意: 题目给出n个点,这些点按照y坐标的升序,若y相同,则按照x的升序顺序输入,问,在这些点中,左下角的点的数量分别在0~n-1的点分别有多少个,写出它们的对应点数.  解题分析: 因为所有点是按照y坐标的升序优先给出,所以后面的点y坐标一定大于等于前面的.于是,判断该点前面有多少点满足在该点的“左下角”,只需看前面的点有几个x坐标小于当前点即可,所以,我们只需要对x坐标建一维树状数组即可求解. #include <cstdio> #include <cs…

题目 题意:按y递增的顺序给出n颗星星的坐标(y相等则x递增),每个星星的等级等于在它左边且在它下边(包括水平和垂直方向)的星星的数量,求出等级为0到n-1的星星分别有多少个. 因为y递增的顺序给出,于是乎这道题跟 y 无关,在每次输入x的时候都去判断比x小,即在x之前的数有多少(ans)个,即星星的等级.然后更新num[ans]++  ( 存ans等级星星的个数num[ans]++ ). 刚开始觉得x不是值吗,怎么一会又变成下标了.其实x一直都是下标,既是题中坐标系的下标,也是树状数组中的下标…

题目: poj 2352 Stars 数星星 题意:已知n个星星的坐标.每个星星都有一个等级,数值等于坐标系内纵坐标和横坐标皆不大于它的星星的个数.星星的坐标按照纵坐标从小到大的顺序给出,纵坐标相同时则按照横坐标从小到大输出. (0 <= x, y <= 32000) 要求输出等级0到n-1之间各等级的星星个数. 分析: 这道题不难想到n平方的算法,即从纵坐标最小的开始搜,每次找它前面横坐标的值比它小的点的个数,两个for循环搞定,但是会超时. 所以需要用一些数据结构去优化,主要是优化找 横坐…