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四色定理

2019-11-14 11:45:31
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四色問題的內(nèi)容是“任何一張地圖只用四種顏色就能使具有共同邊界的國家著上不同的顏色。”也就是說在不引起混淆的情況下一張地圖只需四種顏色來標(biāo)記就行。 From-百度百科

昨天遇到一個(gè)有關(guān)的問題先貼在這里,以后做到新的問題再做總結(jié) Codeforces Round #395 (Div. 2) 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 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

代碼

#include <iostream>#include<cstdio>#include<stdlib.h>using namespace std;int main(){ int n,x1,y1,x2,y2; scanf("%d",&n); printf("YES/n"); for(int i=0;i<n;i++) { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); printf("%d/n",(abs(x1)%2)+1+(abs(y1)%2)*2); } return 0;}
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