POJ2352 Stars[树状数组]

B - Stars

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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

51 15 17 13 35 5

Sample Output

12110

Hint

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

题意:

给你N个星星  找出左下角有多少个星星  然后按照1-n 来分等级 输出每个等级有多少个星星

坐标的输入是根据y从小到大输入的

因为是根据 Y从小到大输入的 X也是从小到大输入的 然后就每次输入的时候 检查x以前的有多少个星星

然后再把自己update进去  注意x是从0开始的  update的时候把x+1

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<string>#include<algorithm>#include<queue>#include<stack>#include<set>#include<map>#include<vector>using namespace std;const int N=32010;int bit[N],num[N];int n;int lowbit(int x){    return x&(-x);}void update(int x,int val){    while (x<=N)    {        bit[x]+=val;        x+=lowbit(x);    }}int query(int x){    int sum=0;    while(x>0)    {        sum+=bit[x];        x-=lowbit(x);    }    return sum;}int main(){    while (~scanf("%d",&n))    {        memset(bit,0,sizeof(bit));        memset(num,0,sizeof(num));        for (int i=0 ; i<n ; i++)        {            int x,y;            scanf("%d%d",&x,&y);            num[query(x+1)]++;            update(x+1,1);//x的最小值从0开始 所以加1        }        for (int i=0 ; i<n ; i++)            printf("%d\n",num[i]);    }    return 0;}