ural1028 Stars

Stars

Time limit: 0.25 second
Memory limit: 64 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.
ural1028 Stars_第1张图片
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 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.

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

input output
5
1 1
5 1
7 1
3 3
5 5
1
2
1
1
0

 

分析:树状数组;

代码:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include <string>
#include <set>
#include 
#include 
#include 
#include 
#include 
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=1e5+10;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,k,t,a[maxn],ans[maxn];
int get(int x)
{
    int sum=0;
    for(int i=x;i;i-=i&(-i))
        sum+=a[i];
    return sum;
}
void add(int x,int y)
{
    for(int i=x;i<=32001;i+=i&(-i))
        a[i]+=y;
}
int main()
{
    int i,j;
    scanf("%d",&n);
    rep(i,1,n)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        x++;
        ans[get(x)]++;
        add(x,1);
    }
    rep(i,0,n-1)printf("%d\n",ans[i]);
    //system("Pause");
    return 0;
}

 

转载于:https://www.cnblogs.com/dyzll/p/5788217.html

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