Ural 1028. Stars 线段树单点更新

1028. Stars

Time limit: 0.25 second
Memory limit: 64 MB

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 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 ofY 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

inputoutput

51 15 17 13 35 5

12110

Problem Author: Pavel Zaletsky 
Problem Source: Ural Collegiate Programming Contest '99

线段树入门级

天下线段树模板出自一家，不愧是notonlysuccess

天文学家时常调查星图，星图在一个平面上表示出所有星星，而且每个星星都有笛卡尔坐标。星星的级别是不在它上面且不在它右面的星星的总数。 天文学家想知道每个星星的级别。 你的任务是计算每个级别的星星总数。

#include<cstdio>#include<cstring>using namespace std;const int MAXN = 15000+5, MAXM = 32000+5;int N, X, Y, level[MAXN];int sum[MAXM<<2];void update(int idx, int L, int R){    int left = (idx<<1), right = (idx<<1)^1;    if (L == X && R == X)        sum[idx]++;    else    {        int mid = ((L+R)>>1);        if (X <= mid)            update(left, L, mid);        else if (mid < X)            update(right, mid+1, R);        sum[idx] = sum[left]+sum[right];    }}int query(int idx, int L, int R, int l, int r){    int left = (idx<<1), right = (idx<<1)^1;    if (L == l && R == r)        return sum[idx];    int mid = ((L+R)>>1);    if (r <= mid)        return query(left, L, mid, l, r);    else if (mid < l)        return query(right, mid+1, R, l, r);    else        return query(left, L, mid, l, mid)+query(right, mid+1, R, mid+1, r);}int main(){    scanf("%d", &N);    for (int i = 1; i <= N; i++)    {        scanf("%d%d", &X, &Y);        level[query(1, 0, 32000, 0, X)]++;        update(1, 0, 32000);    }    for (int i = 0; i < N; i++)        printf("%d\n", level[i]);    return 0;}