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  • Codeforces Round #245 (Div. 2) A. Points and Segments (easy) 贪心

    A. Points and Segments (easy)

    Time Limit: 1 Sec  Memory Limit: 256 MB

    题目连接

    http://codeforces.com/contest/430/problem/A

    Description

    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 integers x1, 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.


    1000000000.

    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 Input

    3 3
    3 7 14
    1 5
    6 10
    11 15

    Sample Output

    0 0 0

    HINT


    题意

    给你一些点,给你一些区间,然后让你染色,让每一个区间红色和黑色的个数都差1

    题解:

      对点排一个序,然后依次染色

    代码:

    //qscqesze
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    #include <map>
    #include <stack>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define maxn 200001
    #define mod 10007
    #define eps 1e-9
    //const int inf=0x7fffffff;   //无限大
    const int inf=0x3f3f3f3f;
    /*
    
    int buf[10];
    inline void write(int i) {
      int p = 0;if(i == 0) p++;
      else while(i) {buf[p++] = i % 10;i /= 10;}
      for(int j = p-1; j >=0; j--) putchar('0' + buf[j]);
      printf("
    ");
    }
    */
    //**************************************************************************************
    inline ll read()
    {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    struct node
    {
        int x,y,z;
    };
    node a[maxn];
    bool cmp(node x,node y)
    {
        return x.x<y.x;
    }
    bool cmp2(node k,node d)
    {
        return k.y<d.y;
    }
    int main()
    {
        int n=read(),m=read();
        for(int i=0;i<n;i++)
        {
            a[i].x=read(),a[i].y=i;
        }
        sort(a,a+n,cmp);
        for(int i=0;i<n;i++)
            if(i%2==0)
                a[i].z=1;
            else
                a[i].z=0;
        sort(a,a+n,cmp2);
        for(int i=0;i<n;i++)
            cout<<a[i].z<<" ";
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4428282.html
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