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

2019-11-14 11:14:01
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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 題意:給n個矩形染色,要求相鄰不能同色。問能否做到。如果能,輸出每個矩形被染得顏色(1,2,3.。。)。 題解:因為四色定理,很明顯可以知道一定是可以的。但是沒有好好理解題干中給出的矩形邊長為奇數的意義。 按大神們的思路來說,因為這個可以把矩形按左下角坐標分成(奇奇,奇偶,偶奇,偶偶)四種類型, 代碼:

#include <bits/stdc++.h>#define ll long longusing namespace std;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;}
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