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Codeforces Round #395 (Div. 2) D

2024-09-10 01:41:22   220人阅读

Description

One of Timofey‘s birthday presents is a colourbook in a shape of an infinite plane. On the plane n rectangles with sides parallel to coordinate axes are situated. All sides of the rectangles have odd length. Rectangles cannot intersect, but they can touch each other.

Help Timofey to color his rectangles in 4 different colors in such a way that every two rectangles touching each other by side would have different color, or determine that it is impossible.

Two rectangles intersect if their intersection has positive area. Two rectangles touch by sides if there is a pair of sides such that their intersection has non-zero length

技术分享
The picture corresponds to the first example
Input

The first line contains single integer n (1?≤?n?≤?5·105) — the number of rectangles.

n lines follow. The i-th of these lines contains four integers x1, y1, x2 and y2 (?-?109?≤?x1?<?x2?≤?109, ?-?109?≤?y1?<?y2?≤?109), that means that points (x1,?y1) and (x2,?y2) are the coordinates of two opposite corners of the i-th rectangle.

It is guaranteed, that all sides of the rectangles have odd lengths and rectangles don‘t intersect each other.

Output

Print "NO" in the only line if it is impossible to color the rectangles in 4 different colors in such a way that every two rectangles touching each other by side would have different color.

Otherwise, print "YES" in the first line. Then print n lines, in the i-th of them print single integer ci (1?≤?ci?≤?4) — the color of i-th rectangle.

Example
input
8
0 0 5 3
2 -1 5 0
-3 -4 2 -1
-1 -1 2 0
-3 0 0 5
5 2 10 3
7 -3 10 2
4 -2 7 -1
output
YES
1
2
2
3
2
2
4
1
题意：问能不能被四种颜色标记地图
解法：四色问题，当然是YES，现在考虑如何染色
我们看左下角如果都是奇数，因为长度是奇数，所以其他的x,y都是偶数，那么其他以奇数为左下角的都不会有交集
嗯，四种颜色嘛。。奇偶排列刚好是四种，于是。。
#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    cout<<"YES"<<endl;
    for(int i=1;i<=n;i++)
    {
        int x,y,a,b;
        cin>>x>>y>>a>>b;
        if(x%2&&y%2)
        {
            cout<<"1"<<endl;
        }
        else if(x%2==0&&y%2)
        {
            cout<<"2"<<endl;
        }
        else if(x%2&&y%2==0)
        {
            cout<<"3"<<endl;
        }
        else if(x%2==0&&y%2==0)
        {
            cout<<"4"<<endl;
        }
    }
    return 0;
}

 

Codeforces Round #395 (Div. 2) D


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