zoukankan      html  css  js  c++  java
  • 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<<" ";
    }
  • 相关阅读:
    求n(n>=2)以内的质数/判断一个数是否质数——方法+细节优化
    poj1185炮兵阵地 正确代码及错误代码分析
    运算符优先级的几点注意
    mod(%)之规律(除数与被除数的正负分析)
    css背景
    Content-Type
    vue数组的增改和v-model的绑定使用Demo
    python open函数关于w+ r+ 读写操作的理解(转)
    http状态码解释
    cookie与token对比(转)
  • 原文地址:https://www.cnblogs.com/qscqesze/p/4428282.html
Copyright © 2011-2022 走看看