Codeforces 763B-Timofey and rectangles

Timofey and rectangles

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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


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：给出n个矩形的左下角和右上角的坐标定点，并保证矩形边长为奇数，给矩形涂色并保证相接触的矩形颜色不同，问能否用四种不同的颜色完成

解题思路：因为所有矩形的边长均为奇数，所以可以根据矩形左下角横纵坐标的奇偶性来把矩形分为四类（奇奇，偶偶，奇偶，偶奇）

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;
const int MAX=100009;

int main()
{
int n;
while(~scanf("%d",&n))
{
printf("YES\n");
int a,b,c,d;
for(int i=0; i<n; i++)
{
scanf("%d%d%d%d",&a,&b,&c,&d);
printf("%d\n",1+2*(abs(a)%2)+abs(b)%2);
}
}
return 0;
}