牛客多校第六场 E Androgynos 自补图

题意：

给定点数，构造自补图，要求输出邻接矩阵，和原图与补图的同构映射。

题解：

只有点数为4k和4k+1的情况才能构造自补图，因为只有这些情况下边数才为偶数。

一种构造方式是，邻接矩阵和同构映射增量构造，每次加四个边xyzw，x和w连接原来的所有边，yz不连，同构映射是交换xy，zw

参考：www.matrix67.com/blog/archives/6221

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int M = 2e3 + ;
const LL mod = ;
const LL lINF = 0x3f3f3f3f3f3f3f3f;
bool mp[M][M];
int t;
int n;
int a[M];
int main()
{
    scanf("%d", &t);
    for (int cnt = ; cnt <= t; cnt++)
    {
        scanf("%d", &n);
        if (n %  == )
        {
            printf("Case #%d: Yes\n", cnt);
            memset(mp, , sizeof mp);
            mp[][] = mp[][] = ;
            mp[][] = mp[][] = ;
            mp[][] = mp[][] = ;
            a[] = ;
            a[] = ;
            a[] = ;
            a[] = ;
            for (int i = ; i <= n; i += )
            {
                mp[i][i + ] = mp[i + ][i] = ;
                mp[i + ][i + ] = mp[i + ][i + ] = ;
                mp[i + ][i + ] = mp[i + ][i + ] = ;
                for (int j = ; j < ; j+=)
                {
                    for (int k = ; k < i; k++)
                    {
                        mp[i + j][k] = mp[k][i + j] = ;
                    }
                }
                a[i] = i + ;
                a[i + ] = i;
                a[i + ] = i + ;
                a[i + ] = i + ;
                swap(a[i+], a[i + ]);
            }
            for (int i = ; i <= n; i++)
            {
                for (int j = ; j <= n; j++)
                {
                    printf("%d", mp[i][j]);
                }
                puts("");
            }
            for (int i = ; i < n; i++)
            {
                printf("%d ", a[i]);
            }
            printf("%d\n", a[n]);
        }
        else if (n %  == )
        {
            printf("Case #%d: Yes\n", cnt);
            memset(mp, , sizeof mp);
            a[] = ;
            for (int i = ; i <= n; i += )
            {
                mp[i][i + ] = mp[i + ][i] = ;
                mp[i + ][i + ] = mp[i + ][i + ] = ;
                mp[i + ][i + ] = mp[i + ][i + ] = ;
                for (int j = ; j < ; j+=)
                {
                    for (int k = ; k < i; k++)
                    {
                        mp[i + j][k] = mp[k][i + j] = ;
                    }
                }
                a[i] = i + ;
                a[i + ] = i;
                a[i + ] = i + ;
                a[i + ] = i + ;
                swap(a[i+], a[i + ]);
            }
            for (int i = ; i <= n; i++)
            {
                for (int j = ; j <= n; j++)
                {
                    printf("%d", mp[i][j]);
                }
                puts("");
            }
            for (int i = ; i < n; i++)
            {
                printf("%d ", a[i]);
            }
            printf("%d\n", a[n]);
        }
        else
        {
            printf("Case #%d: No\n", cnt);
        }
    }
}