CF323B.Tournament-graph

In this problem you have to build tournament graph, consisting of n vertices, such, that for any oriented pair of vertices (v, u) (v ≠ u) there exists a path from vertex v to vertex u consisting of no more then two edges.

A directed graph without self-loops is a tournament, if there is exactly one edge between any two distinct vertices (in one out of two possible directions).

本题要求构造一个包含 $n$ 个顶点的竞赛图，使得对任意一对有序顶点 $(v,\,u)$（其中 $v \ne u$），均存在一条从顶点 $v$ 到顶点 $u$ 的有向路径，且该路径所含边数不超过两条。

一个无自环的有向图称为竞赛图，当且仅当对任意两个不同的顶点，二者之间恰好存在一条有向边（即两个可能方向中仅取其一）。

The first line contains an integer n (3 ≤ n ≤ 1000), the number of the graph's vertices.

第一行包含一个整数 $n$（$3 \leq n \leq 1000$），表示图的顶点数。

Print -1 if there is no graph, satisfying the described conditions.

Otherwise, print n lines with n integers in each. The numbers should be separated with spaces. That is adjacency matrix a of the found tournament. Consider the graph vertices to be numbered with integers from 1 to n. Then a__v, u = 0, if there is no edge from v to u, and a__v, u = 1 if there is one.

As the output graph has to be a tournament, following equalities must be satisfied:

• a__v, u + a__u, v = 1 for each v, u (1 ≤ v, u ≤ n; v ≠ u);
• a__v, v = 0 for each v (1 ≤ v ≤ n).

如果不存在满足所述条件的图，则输出 -1。

否则，输出 n 行，每行包含 n 个整数，各数之间用空格分隔。即输出所找到的竞赛图（tournament）的邻接矩阵 a。设图的顶点编号为 1 到 n 的整数，则当不存在从顶点 v 指向顶点 u 的有向边时，a_{v,u} = 0；当存在该有向边时，a_{v,u} = 1。

由于输出图必须是一个竞赛图，以下等式必须成立：

• 对任意 v, u（1 ≤ v, u ≤ n 且 v ≠ u），有 a_{v,u} + a_{u,v} = 1；
• 对任意 v（1 ≤ v ≤ n），有 a_{v,v} = 0。