POJ 2352-Stars（树状数组-星系）-蒲公英云

POJ 2352-Stars（树状数组-星系）

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 46418		Accepted: 20020

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.

Input

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.

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

Sample Output

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

Source

Ural Collegiate Programming Contest 1999

题目意思：

给出N个星星的坐标，在45°方向上能连成一条线的星星为一个星系的话，如果有M颗星星连成一线则最右上方星星的level是M-1。

例如图中level=0的星星是1，level=1的星星是2和4，level=2的星星是3，level=3的星星是5。

解题思路：

因为y坐标顺序给定，所以按输入处理x坐标，每次计算并记录其左边的星星的个数，然后对其右边的所有点更新增一。

注意x可能为0，而树状数组下标从1 开始。

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<map>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define INF 0xfffffff
#define MAXN 32005
int n;
int c[MAXN],cnt[MAXN];
int lowbit(int x)
{
    return x&(-x);
}
void update(int i,int x)
{
    while(i<MAXN)
    {
        c[i]+=x;
        i+=lowbit(i);
    }
}
int sum(int x)
{
    int i,res=0;//res是权值和
    for(i=x; i>0; i-=lowbit(i))
        res+=c[i];
    return res;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("G:/cbx/read.txt","r",stdin);
    //freopen("G:/cbx/out.txt","w",stdout);
#endif
    while(~scanf("%d",&n))
    {
        memset(c,0,sizeof(c));
        memset(cnt,0,sizeof(cnt));
        for(int i=0; i<n; i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            ++x;//避免横坐标为0的情况
            int t=sum(x);
            ++cnt[t];
            update(x,1);//c[x]之后的全部+1
        }
        for(int i=0; i<n; i++)//输出
            printf("%d\n",cnt[i]);
    }
    return 0;
}