牛客多校6 G.Grid Coloring【构造】

做法：在注意到n==1,k==1,和无法让所有边均分颜色这几种情况无解后，因为横竖边数量的奇偶性不同，所以直接一排横一排竖地按（1~k）的顺序给每条边标号（染色）就能保证条件123同时成立

CODE

#include <bits/stdc++.h>
#define dbug(x) cout << #x << "=" << x << endl
#define eps 1e-8
#define pi acos(-1.0)
 
using namespace std;
typedef long long LL;
 
const int inf = 0x3f3f3f3f;
 
template<class T>inline void read(T &res)
{
   char c;T flag=1;
   while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
   while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
}

const int maxn = 207;

int t, n, k;

vector<int> l[maxn], r[maxn];

int main()
{
    read(t);
    while(t--) {
        read(n); read(k);
        if(k == 1 || n == 1) {
            puts("-1");
            continue;
        }
        int num = (n * (n + 1)) * 2;
        if(num % k) {
            puts("-1");
            continue;
        }
        for ( int i = 1; i <= n + 1; ++i ) {
            l[i].clear();
            r[i].clear();
        }
        int kk = 1;
        int L = 0;
        for ( int i = 1; i <= n + n + 1; ++i ) {
            if(i % 2) {
                ++L;
                for ( int j = 1; j <= n; ++j ) {
                    l[L].push_back(kk);
                    kk++;
                    if(kk > k) {
                        kk = 1;
                    }
                }
            }
            else {
                for ( int j = 1; j <= n + 1; ++j ) {
                    r[j].push_back(kk);
                    kk++;
                    if(kk > k ) {
                        kk = 1;
                    }
                }
            }   
        }
        for ( int i = 1; i <= L; ++i ) {
            for(auto it : l[i]) {
                cout << it << ' ';
            }
            puts("");
        }
        for ( int i = 1; i <= n + 1; ++i ) {
            for(auto it : r[i]) {
                cout << it << ' ';
            }
            puts("");
        }
    }
    return 0;
}