Codeforces Round #395 (Div. 2) D. Timofey and rectangles(思路)
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题目地址:http://codeforces.com/contest/764/problem/D
思路:考虑两相接触的矩形,由于矩形边长为奇数,所以若能相接触必然接触的两个矩形横纵坐标(以某一顶点代表此矩形)奇偶性至少有一个不同(左右:横坐标奇偶性不同;上下:纵坐标奇偶性不同)。而坐标共有四种:奇奇,偶偶,奇偶,偶奇,颜色有四种,保证可接触的矩形颜色不同即可:奇奇4,奇偶3,偶奇2,偶偶1。
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int main(){ int n; scanf("%d",&n); printf("YES\n"); for(int i=0;i<n;i++) { int a,b,c,d; 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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