poj 2352 Stars

Stars

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 33003 Accepted: 14422

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.

由于y坐标是升序的且坐标不重复，所以在星星A后面输入的星星的x,y坐标不可能都小于等于星星A。假如当前输入的星星为（3,3），易得我们只需要去找树状数组中小于等于3的值就可以了，即Getsum(3)。注意：A[i]表示x坐标为i的个数，C[]为A[]的树状数组，那么GetSum(i)就是序列中前i个元素的和，即x小于等于i的星星数。

本题还是一点要注意：星星坐标的输入可以是（0,0），所以我们把x坐标统一加1，(加1 后不影响所有所以数据间大小关系)然后用树状数组实现。

#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define inf -0x3f3f3f3f#define FOR(a,b) for(int i =  a ; i < b ; i++)#define mem0(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define FOPENIN(IN) freopen(IN, "r", stdin)#define FOPENOUT(OUT) freopen(OUT, "w", stdout)const  int NUM =35000;int lowbit(int x){    return x&(-x);}int N,lev[NUM],c[NUM];int Getsum(int x){    int ans = 0 ;    while(x) {        ans += c[x];        x -= lowbit(x);    }    return ans;}void add(int x){    while(x <= NUM){        c[x]+=1 ;        x += lowbit(x);    }}int main(){    while(scanf("%d",&N)!=EOF){        mem0(c);        int x,y;        for(int i = 0 ; i < N ; i++){            scanf("%d %d",&x,&y);            x++;            lev[Getsum(x)]++;//lev用来记录等级为i的个数            add(x);        }        for(int i = 0 ;i < N ; i++)        printf("%d\n",lev[i]);    }    return 0;}