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Stars （poj 2352 樹狀數組）

編輯：C++入門知識

Stars （poj 2352 樹狀數組）

Language: Stars Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 35169 Accepted: 15259

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

題意：有n個星星的坐標，如果某個星星坐標為(x, y), 它的左下位置為：(x0,y0)，x0<=x 且y0<=y。如果左下位置有a個星星，就表示這個星星屬於level x
按照y遞增，如果y相同則x遞增的順序給出n個星星，求出所有level水平的數量。

思路：最典型的樹狀數組，第一次做。。

代碼：

#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 32005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

int bit[maxn];
int ans[maxn/2];

int lowbit(int k)
{
    return k&(-k);
}

int sum(int i)
{
    int s=0;
    while (i>0)
    {
        s+=bit[i];
        i-=lowbit(i);
    }
    return s;
}

void add(int i,int x)
{
    while (i