Codeforces 430 A. Points and Segments (easy)

最新推荐文章于 2020-08-30 11:49:02 发布

原创最新推荐文章于 2020-08-30 11:49:02 发布 · 1.5k 阅读

0 ·

CC 4.0 BY-SA版权

水题专栏收录该内容

126 篇文章

订阅专栏

本篇介绍了一个几何问题：如何在OX轴上绘制n个不同颜色的点和m条线段，使得每条线段内的红蓝点数量之差不超过1。提供了输入输出格式及示例，并附带了解决该问题的C++代码。

Iahub isn't well prepared on geometry problems, but he heard that this year there will be a lot of geometry problems on the IOI selection camp. Scared, Iahub locked himself in the basement and started thinking of new problems of this kind. One of them is the following.

Iahub wants to draw n distinct points and m segments on the OX axis. He can draw each point with either red or blue. The drawing is good if and only if the following requirement is met: for each segment [li, ri] consider all the red points belong to it (ri points), and all the blue points belong to it (bi points); each segment i should satisfy the inequality |ri - bi| ≤ 1.

Iahub thinks that point x belongs to segment [l, r], if inequality l ≤ x ≤ r holds.

Iahub gives to you all coordinates of points and segments. Please, help him to find any good drawing.

Input

The first line of input contains two integers: n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 100). The next line contains n space-separated integersx1, x2, ..., xn (0 ≤ xi ≤ 100) — the coordinates of the points. The following m lines contain the descriptions of the m segments. Each line contains two integers li and ri (0 ≤ li ≤ ri ≤ 100) — the borders of the i-th segment.

It's guaranteed that all the points are distinct.

Output

If there is no good drawing for a given test, output a single integer -1. Otherwise output n integers, each integer must be 0 or 1. The i-th number denotes the color of the i-th point (0 is red, and 1 is blue).

If there are multiple good drawings you can output any of them.

Sample test(s)

input
3
7 14
5
10
15

output
0 0

input
4
2 3
2
3
6
2

output
0 1 

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

using namespace std;

int n,m;
struct POINT
{
    int x,c,id;
}point[110];

bool cmp1(POINT a,POINT b)
{
    return a.x<b.x;
}

bool cmp2(POINT a,POINT b)
{
    return a.id<b.id;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&point[i].x);
        point[i].id=i;
    }
    for(int i=0;i<m;i++) scanf("%*d%*d");
    sort(point,point+n,cmp1);
    for(int i=0;i<n;i++)
    {
        point[i].c=i%2;
    }
    sort(point,point+n,cmp2);
    for(int i=0;i<n;i++)
    {
        printf("%d ",point[i].c);
    }
    putchar(10);
    return 0;
}