D. Timofey and rectangles

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

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.

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.

80 0 5 32 -1 5 0-3 -4 2 -1-1 -1 2 0-3 0 0 55 2 10 37 -3 10 24 -2 7 -1output
YES12232241
这个题根据他一直强调的边长为奇数的重要信息，画图来看：
相邻的4个区域正好对应着左下角的点的坐标的奇偶性。
分别是奇奇，奇偶，偶奇，偶偶。任何相邻的4个都必定是一样的规律。
所以我们根据这4种情况来编号就行了。
#include <bits/stdc++.h>using namespace std;const int MAXN=2e5+7;const int inf =0x3f3f3f3f;int n,m;int main(){    int i;    scanf("%d",&n);    puts("YES");    int x,y;    for(i=0;i<n;++i)    {        scanf("%d%d%*d%*d",&x,&y);        printf("%d/n",(x&1)*2+(y&1)+1);    }    return 0;}