(简单) POJ 2352 Stars，Treap。-白红宇

(简单) POJ 2352 Stars，Treap。

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发布时间：2019-06-07

本文共 2388 字，大约阅读时间需要 7 分钟。

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.

　　题目就是求每个星星的等级，然后统计某个等级的有几个星星，等级的话就是左下边的星星的个数。

　　把星星排序，从左到右，一次统计每个星星，把在他左边的加入二叉树，然后查找当前星星下面的就好了。

代码如下：

// ━━━━━━神兽出没━━━━━━//      ┏┓       ┏┓//     ┏┛┻━━━━━━━┛┻┓//     ┃           ┃//     ┃     ━     ┃//     ████━████   ┃//     ┃           ┃//     ┃    ┻      ┃//     ┃           ┃//     ┗━┓       ┏━┛//       ┃       ┃//       ┃       ┃//       ┃       ┗━━━┓//       ┃           ┣┓//       ┃           ┏┛//       ┗┓┓┏━━━━━┳┓┏┛//        ┃┫┫     ┃┫┫//        ┗┻┛     ┗┻┛//// ━━━━━━感觉萌萌哒━━━━━━// Author        : WhyWhy// Created Time  : 2015年07月17日 星期五 14时44分13秒// File Name     : 1195.cpp#include 
       
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                   using namespace std;const int MaxN=1100;int C[MaxN][MaxN];int N;inline int lowbit(int x){ return x&(-x);}void add(int x,int y,int d){ int t; while(x<=N) { t=y; while(t<=N) { C[x][t]+=d; t+=lowbit(t); } x+=lowbit(x); }}int query(int x,int y){ int ret=0; int t; while(x>0) { t=y; while(t>0) { ret+=C[x][t]; t-=lowbit(t); } x-=lowbit(x); } return ret;}int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int a,b,c,d,e; while(1) { scanf("%d",&a); if(a==1) { scanf("%d %d %d",&b,&c,&d); add(b+1,c+1,d); } else if(a==2) { scanf("%d %d %d %d",&b,&c,&d,&e); printf("%d\n",query(d+1,e+1)-query(d+1,c)-query(b,e+1)+query(b,c)); } else if(a==0) { scanf("%d",&N); memset(C,0,sizeof(C)); } else break; } return 0;}

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转载于:https://www.cnblogs.com/whywhy/p/4654672.html

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