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

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 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.

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.

80 0 5 32 -1 5 0-3 -4 2 -1-1 -1 2 0-3 0 0 55 2 10 37 -3 10 24 -2 7 -1output
YES12232241
解题思路：
因为矩形的边为奇数长度 根据四色定理，染色一定会成功。
(1)我们只看左下角坐标，如果两个数值都为奇数，那么右上角坐标一定两个都为偶数，所以所有左下标坐标为奇数的不会相交，可赋值为1。
(2) 如果x轴为偶数，可能与1的情况左右相邻赋值为2。
(3) 如果y轴为偶数，可能与1的情况左右相邻，赋值为3。
(4) 其余赋值为4。
时间复杂度：O(n) 空间复杂度：O(n)
#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <iomanip>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}typedef  vector<int> vi;int dir[4][2]={{1,0},{0,-1},{-1,0},{0,1}};const int N = 1e5+5;int n;int main(){    int a,b,c,d;    cin >> n;    cout << "YES"<<endl;     while(n--)    {        cin >> a>> b>>c>>d;        if(a%2&&b%2)        {            cout << 1<<endl;        }        else if(a%2==0&&b%2)        {            cout << 2<<endl;        }        else if(a%2&&b%2==0)        {            cout << 3<<endl;        }        else        {            cout << 4<<endl;        }    }    return 0;   } 膜一发dls
#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <map>#include <set>#include <cassert>using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define pb push_back#define mp make_pair#define all(x) (x).begin(),(x).end()#define fi first#define se second#define SZ(x) ((int)(x).size())typedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;const ll mod=1000000007;ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}// headint n,x,y;int main() {	scanf("%d",&n);	puts("YES");	rep(i,0,n) {		scanf("%d%d%*d%*d",&x,&y);		printf("%d/n",2*(x&1)+(y&1)+1);	}}