Codeforces Round #395 (Div. 2)-D. Timofey and rectangles

原题链接

D. 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

x1, y1奇偶性相同的矩形不会向接触

#include <bits/stdc++.h> #define maxn 200005 using namespace std; typedef long long ll; int main(){ // freopen("in.txt", "r", stdin); int n, x1, y1, x2, y2; scanf("%d", &n); puts("YES"); while(n--){ scanf("%d%d%d%d", &x1, &y1, &x2, &y2); x1 = abs(x1); y1 = abs(y1); if(x1 % 2 == 1 && y1 % 2 == 1) puts("1"); else if(x1 % 2 == 0 && y1 % 2 == 1) puts("2"); else if(x1 % 2 == 1 && y1 % 2 == 0) puts("3"); else puts("4"); } return 0; }