Stars(树状数组)

传送门Stars

题目描述

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.

输入

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.

输出

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.

样例

  • Input
    5
    1 1
    5 1
    7 1
    3 3
    5 5
  • Output
    1
    2
    1
    1
    0

题解

  • 题意就是每个星星有一个坐标,如果在一颗星星左下角有k个星星,那么这颗星星就是第k级。现在给定n个星星的坐标,且给定的条件中星星的y坐标递增,求出0到n-1每一级的星星的个数。
  • 对于当前的星星,统计前面x比它小的星星个数,就是他的级数
  • 由于时间复杂度的关系,这里要用到树状数组。用star[x]保存横坐标覅为x的星星的出现次数,对于每个星星,求一次前缀和即可。

Code

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#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#define LL long long int
#define INIT(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,b,a) for(int i=b-1;i>=a;i--)
//b——0,-1,128,0x3f,127 ,字符
const double Pi = acos(-1);
const double E = exp(1.0);
const LL mod =1e9+7;
const int MAX=0x7fffffff;
const int MIN=-0x7fffffff;
const int INF=0x3f3f3f3f;
using namespace std;
int n;int tree[40000];int num[40000];
int lowbit(int x){
return x & -x;
}
void add(int x,int p){
while(x<40000){
tree[x]+=p;
x+=lowbit(x);
}
}
int sum(int x){
int t=0;
while(x>0){
t+=tree[x];
x-=lowbit(x);
}
return t;
}
int main()
{
while(~scanf("%d",&n)){
INIT(tree,0);INIT(num,0);
int x,y;
for(int i=1;i<=n;i++){
scanf("%d %d",&x,&y);
x++;
num[sum(x)]++;
add(x,1);
}
for(int i=0;i<n;i++)
printf("%d\n",num[i]);
}
return 0;
}
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