Codeforces Round #395 (Div. 2)D

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根据四色原理我们可以确定一定是YES

主要是根据边长一定是奇数这句话，对于矩形左下角的点，如果两个矩形同色，那么两个点的横纵坐标的奇偶性一定相同，这样就是四种情况，四种颜色

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>#include <vector>#include <map>#include <cmath>#include <stdlib.h>using namespace std;const double PI = acos(-1.0);const double eps = 0.1;const int MAX = 5e5+10;const int mod = 1e9+7;const double pp = 1000000000.0+10;struct node{    int x1,x2,y1,y2;} lxt[MAX];int n;int main(){    cin>>n;    for(int i = 0; i<n; ++i)        cin>>lxt[i].x1>>lxt[i].y1>>lxt[i].x2>>lxt[i].y2;    cout<<"YES"<<endl;    for(int i = 0; i<n; ++i)    {        if(lxt[i].x1<0)lxt[i].x1*=-1;        if(lxt[i].y1<0)lxt[i].y1*=-1;        if(lxt[i].x1%2==0&&lxt[i].y1%2==0)cout<<1<<endl;        else if(lxt[i].x1%2&&lxt[i].y1%2==0)cout<<2<<endl;        else if(lxt[i].x1%2==0&&lxt[i].y1%2)cout<<3<<endl;        else if(lxt[i].x1%2&&lxt[i].y1%2)cout<<4<<endl;    }    return 0;}

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