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POJ2352_Stars(段树/单点更新)

时间:2015-09-12 21:43:17      阅读:127      评论:0      收藏:0      [点我收藏+]

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解决报告

意甲冠军:

坐标。查找在数星星的左下角每颗星星。

思考:

横轴作为间隔,已知的输入是所述第一到y排序再次x次序。每次添加一个点来查询点x多少分离开坐标,然后更新点。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
int sum[200000];
struct node {
    int x,y;
} p[20000];
void push_up(int root) {
    sum[root]=sum[root*2]+sum[root*2+1];
}
void update(int root,int l,int r,int p,int v) {
    int mid=(l+r)/2;
    if(l==r) {
        sum[root]++;
        return;
    }
    if(p<=mid)update(root*2,l,mid,p,v);
    else update(root*2+1,mid+1,r,p,v);
    push_up(root);
}
int q_sum(int root,int l,int r,int ql,int qr) {
    if(ql>r||qr<l)return 0;
    if(ql<=l&&r<=qr)return sum[root];
    int mid=(l+r)/2;
    return q_sum(root*2,l,mid,ql,qr)+q_sum(root*2+1,mid+1,r,ql,qr);
}
int main() {
    int n,i,j,m=32000;
    int _hash[20000];
    scanf("%d",&n);
    memset(_hash,0,sizeof(_hash));
    for(i=0; i<n; i++) {
        scanf("%d%d",&p[i].x,&p[i].y);
        _hash[q_sum(1,0,m,0,p[i].x)]++;
        update(1,0,m,p[i].x,1);
    }
    for(i=0; i<n; i++)
        printf("%d\n",_hash[i]);
    return 0;
}

Stars
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 32329   Accepted: 14119

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. 
技术分享

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.

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POJ2352_Stars(段树/单点更新)

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原文地址:http://www.cnblogs.com/lcchuguo/p/4803522.html

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