意甲冠军  中国

依据Havel-Hakimi定理构图即可咯  先把顶点按度数从大到小排序  可图的话  度数大的顶点与它后面的度数个顶点相连肯定是满足的  出现了-1就说明不可图了

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

const int N = 20;

int mat[N][N], ord[N];

bool cmp(int i, int j)

{

    return mat[i][0] > mat[j][0];

}

int main()

{

    int cas, i, j, k, t, n;

    scanf("%d", &cas);

    while(cas--)

    {

        memset(mat, 0, sizeof(mat));

        scanf("%d", &n);

        for(i = 1; i <= n; ++i)

        {

            scanf("%d", &mat[i][0]);

            ord[i] = i;

        }

        for(i = 1; i <= n; ++i)

        {

            sort(ord + i, ord + n + 1, cmp);

            t = ord[i];

            if(mat[t][0] < 0) break;

            for(j = 1; j <= mat[t][0]; ++j)

            {

                k = ord[i + j];

                mat[t][k] = mat[k][t] = 1;

                --mat[k][0];

            }

        }

        if(i <= n) printf("NO\n");

        else

        {

            printf("YES\n");

            for(i = 1; i <= n; ++i)

            {

                for(int j = 1; j <= n; ++j)

                    printf("%d ", mat[i][j]);

                printf("\n");

            }

        }

        if(cas) printf("\n");

    }

    return 0;

}

Description

Input

Output

Sample Input

3

7

4 3 1 5 4 2 1

6

4 3 1 4 2 0

6

2 3 1 1 2 1

Sample Output

YES

0 1 0 1 1 0 1

1 0 0 1 1 0 0

0 0 0 1 0 0 0

1 1 1 0 1 1 0

1 1 0 1 0 1 0

0 0 0 1 1 0 0

1 0 0 0 0 0 0 

NO

YES

0 1 0 0 1 0

1 0 0 1 1 0

0 0 0 0 0 1

0 1 0 0 0 0

1 1 0 0 0 0

0 0 1 0 0 0

Source