【Codeforces Round395】 Codeforces 764D 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 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.

如果两个矩形接触，在接触的那一维上坐标的奇偶性一定不同。这样恰好根据两个坐标的奇偶性分成四类。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){    int n,x,y;    printf("YES\n");    scanf("%d",&n);    while (n--)    {        scanf("%d%d%*d%*d",&x,&y);        printf("%d\n",2*(x&1)+(y&1)+1);    }}