poj 2352（树状数组）

Stars
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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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
5
1 1
5 1
7 1
3 3
5 5
Sample Output
1
2
1
1
0

题意：给你星星的坐标，顺序是 y 坐标非递减，x 坐标严格递增；星星的等级就是它左边和正下方（x < X || y <= Y )星星的个数，有多少个就是几级，最后问你等级在0—>n-1 的星星各有多少个，每个等级的星星个数对应一行；
题解：由于给出星星的顺序已知，所以用树状数组先查询给出的 x 坐标的星星有多少个就是几级，然后再把这个坐标压入树状数组；（注意：树状数组只能从 1 开始，所以每个 x 要加 1）；

#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
#define M 32010
#define RCL(a, b) memset(a, b, sizeof(a))
vector<int> vec[M];
int level[M], tree[M];
int lowbit(int x)
{
    return x & (-x);
}
void addtree(int x)
{
    while(x < M)
    {
        tree[x]++;
        x += lowbit(x);
    }
}
int query(int x)
{
    int sum = 0;
    while(x > 0)
    {
        sum += tree[x];
        x -= lowbit(x);
    }
    return sum;
}

int main()
{
    int n, x, y, vis;
    while(scanf("%d", &n) != EOF)
    {
        RCL(tree, 0);
        RCL(level, 0);
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d", &x, &y);
            level[query(x+1)]++;//用 leave 数组记录等级个数
            addtree(x+1);//压入树状数组
        }
        for(int i=0; i<n; i++)
        {
            printf("%d\n", level[i]);
        }
    }

    return 0;
}