poj 2352 一维树状数组

题目来源Ural Collegiate Programming Contest 1999 poj上的链接点这儿（这个是难得的1Y"o((>ω< ))o" ）

下面放出题目 

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29948		Accepted: 13088

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

5
1 1
5 1
7 1
3 3
5 5

Sample Output

1
2
1
1
0

Hint

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

题目就是给出了好多点的二维平面坐标，然后用level来表示横纵坐标都不大于这个点坐标的点的个数。比如样例中(1,1)的level是0，(5,5)的level是3。然后要依次输出leve=0, 1, 2 ……, n - 1的点的个数，样例中level = 0的点有一个， = 1的有两个，=2 和 3的各有一个，所以输出时1 2 1 1 0.

我们可以将点按照坐标序（先比较横坐标，横坐标相等时比较纵坐标）排一遍，这时，对每个点，我们只需统计已经出现过的点中，纵坐标不大于它纵坐标的点的个数。我们用sum[i]来记录已经出现过的纵坐标为i的点的个数。但是由于数据量较大，如果每次都要累加的话，肯定会TLE，所以我们需要一个数据结构来快速求和，很明显，树状数组便符合这个要求（复杂点的线段树当然也可以）（因为是树状数组，所以sum[]数组就改成了bintree[]数组）。

而且，由于初始给出的数据是按照纵坐标递增的顺序给出的，我们可以再开始便对纵坐标进行离散化处理，这样就避免了纵坐标数据范围的问题。

下面放出代码。

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)>(b)?(b):(a))
#define rep(i,initial_n,end_n) for(int (i)=(initial_n);(i)<(end_n);i++)
#define repp(i,initial_n,end_n) for(int (i)=(initial_n);(i)<=(end_n);(i)++)
#define reep(i,initial_n,end_n) for((i)=(initial_n);(i)<(end_n);i++)
#define reepp(i,initial_n,end_n) for((i)=(initial_n);(i)<=(end_n);(i)++)
#define eps 1.0e-9
#define MAX_N

using namespace std;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef __int64 ll;
typedef unsigned __int64 ull;

int bintree[15010], level[15010];

struct node {
int x, y;
friend bool operator < (node a, node b) {
if(a.x < b.x) return true;
else if(a.x == b.x && a.y < b.y) return true;
else return false;
}
};

node a[15010];

void init() {
memset(bintree, 0, sizeof(bintree));
memset(level, 0, sizeof(level));
}

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

void update(int pos, int val, int len) {
while(pos <= len) bintree[pos] += val, pos += lowbit(pos);
}

int getsum(int pos) {
int sum = 0;
while(pos > 0) sum+= bintree[pos], pos -= lowbit(pos);
return sum;
}

int main() {
int n;
scanf("%d", &n);
rep(i, 0, n) scanf("%d%d", &a[i].x, &a[i].y);
int tmp;
tmp = a[0].y, a[0].y = 1;
repp(i, 1, n) {
if(a[i].y == tmp) a[i].y = a[i-1].y;
else tmp = a[i].y, a[i].y = a[i-1].y + 1;
}
sort(a, a + n);
init();
rep(i, 0, n) {
int sum = 0;
sum = getsum(a[i].y);
level[sum]++;
update(a[i].y, 1, n);
}
rep(i, 0, n) printf("%d\n", level[i]);
return 0;
}