AtCoder Regular Contest 089 C - Traveling

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Problem StatementAtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0,0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i.

If AtCoDeer is at point (x,y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x−1,y), (x,y+1) and (x,y−1). Note that he cannot stay at his place. Determine whether he can carry out his plan.

Constraints

• 1 ≤ N ≤ 10⁵
• 0 ≤ x_i ≤ 10⁵
• 0 ≤ y_i ≤ 10⁵
• 1 ≤ t_i ≤ 10⁵
• t_i < t_i+1 (1 ≤ i ≤ N−1)
• All input values are integers.

InputInput is given from Standard Input in the following format:

N
t1 x1 y1
t2 x2 y2
:
tN xN yN

Output

If AtCoDeer can carry out his plan, print Yes; if he cannot, print No.

Sample Input 1

2
3 1 2
6 1 1

Sample Output 1

Yes

For example, he can travel as follows: (0,0), (0,1), (1,1), (1,2), (1,1), (1,0), then (1,1).

Sample Input 2

1
2 100 100

Sample Output 2

No

It is impossible to be at (100,100) two seconds after being at (0,0).

Sample Input 3

2
5 1 1
100 1 1

Sample Output 3

No

这个方法真的很奇淫：

1. 两点之间的步数小于需要的步数，则不能到达（可行性剪枝）

2. 到达目标点的步数非偶数步，否则不能到达（奇偶剪枝）

3. 否则便可以到达

根本不需要写DFS

#include <iostream>
#include <cmath>

using namespace std;
int main(){
	int n, tt, tx, ty; 
	int t = 0, x = 0, y = 0, flag = 0;
	cin >> n;
	for (int i = 0; i<n; i++){
		cin >> tt >> tx >> ty;
		if (abs(tx - x) + abs(ty - y)>(tt - t) || (abs(tx - x) + abs(ty - y)) % 2 != (tt - t) % 2)
			flag = 1;
		t = tt;
		x = tx;
		y = ty;
	}
	if (!flag)	cout << "Yes" << endl;
	else	cout << "No" << endl;

	return 0;
}