AtCoder Regular Contest 089 C - Traveling

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
(xi,yi)
at time ti.
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
≤
xi
≤
105
0
≤
yi
≤
105
1
≤
ti
≤
105
ti
<
ti+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;
}