Discription
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 vertexv 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).
Input
The first line contains an integer n (3 ≤ n ≤ 1000), the number of the graph's vertices.
Output
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 av, u = 0, if there is no edge from v to u, and av, u = 1 if there is one.
As the output graph has to be a tournament, following equalities must be satisfied:
- av, u + au, v = 1 for each v, u (1 ≤ v, u ≤ n; v ≠ u);
- av, v = 0 for each v (1 ≤ v ≤ n).
Example
3
0 1 0
0 0 1
1 0 0
4
-1
构造题,比较迷。
首先我们如果有了n个点的合法图是很容易构造出n+2个点的合法图的。
首先可以把1的入点看成一类点,1的出点看成一类点,然后再加上1本身和n+1和n+2,我们现在就有了5个点。
而我们的目的是让任意一对点都在至少一个三元环 (形如a->b,b->c,c->a) 出现。
所以直接xjb构造就行了,这个其实可以不用手算,直接写个程序跑一跑也是可以的2333(我是不会告诉你们我的连边方案就是电脑枚举出来的哈哈)
但是发现4没有答案,导致我一开始以为偶数都是gg的然后就WA了。。。
后来写了个搜索发现6是有答案的2333,所以是偶数的话特判完了之后直接从6的图往后跑就行了,n=6的图可以搜索也可以手玩(反正写搜索就当练暴力了2333)
#include<bits/stdc++.h> #define ll long long #define maxn 1005 using namespace std; int a[maxn][maxn]; int n,m,k,tone[maxn]; int main(){ scanf("%d",&n); if(!(n&1)){ if(n<6){ puts("-1"); return 0; } a[1][5]=a[1][6]=1; a[2][1]=a[2][6]=1; a[3][1]=a[3][2]=a[3][5]=1; a[4][1]=a[4][2]=a[4][3]=1; a[5][2]=a[5][4]=1; a[6][3]=a[6][4]=a[6][5]=1; tone[2]=tone[3]=tone[4]=1; for(int i=7;i<=n;i+=2){ tone[i]=1; a[i][1]=1,a[1][i+1]=1,a[i+1][i]=1; for(int j=2;j<i;j++){ if(tone[j]) a[j][i]=1,a[i+1][j]=1; else a[j][i+1]=1,a[i][j]=1; } } } else{ a[1][2]=a[2][3]=a[3][1]=1; tone[3]=1; for(int i=4;i<=n;i+=2){ tone[i]=1; a[i][1]=1,a[1][i+1]=1,a[i+1][i]=1; for(int j=2;j<i;j++){ if(tone[j]) a[j][i]=1,a[i+1][j]=1; else a[j][i+1]=1,a[i][j]=1; } } } for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++) printf("%d ",a[i][j]); puts(""); } /* for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) a[i][j]=(a[i][j]?1:1<<30); for(int i=1;i<=n;i++) a[i][i]=0; for(int k=1;k<=n;k++) for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(a[i][k]+a[k][j]<a[i][j]) a[i][j]=a[i][k]+a[k][j]; for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(a[i][j]>2) puts("No"); */ return 0; }