杭电1498（50 years, 50 colors）

Input

There will be multiple input cases.Each test case begins with two integers n, k. n is the number of rows and columns of the balloons (1 <= n <= 100), and k is the times that ginving to each student(0 < k <= n).Follow a matrix A of n*n, where Aij denote the color of the ballon in the i row, j column.Input ends with n = k = 0.

 

Output

For each test case, print in ascending order all the colors of which are impossible to be crashed by a student in k times. If there is no choice, print "-1".

 

Sample Input

1 1

1

2 1

1 1

1 2

2 1

1 2

2 2

5 4

1 2 3 4 5

2 3 4 5 1

3 4 5 1 2

4 5 1 2 3

5 1 2 3 4

3 3

50 50 50

50 50 50

50 50 50

0 0

 

Sample Output

-1

1

2

1 2 3 4 5

-1

#include<stdio.h>
#include<string.h>
int n,m;
int map[110][110],a[110],b[110];
int count[100],f[100];
int find(int x,int y)
{
    int i;
    for(i=1;i<=n;i++)
    {
        if(map[x][i]==y&&b[i]==0)//颜色要一致，且没有覆盖过
        {
            b[i]=1;//表示此次覆盖
             if(a[i]==0||find(a[i],y))
             {
                 a[i]=x;//形成新的交替
                 return 1;
             }
        }
    }
    return 0;
}
int met(int x)
{
    int i,sum;
    sum=0;
    memset(a,0,sizeof(a));//每一种颜色a都要初始化
    for(i=1;i<=n;i++)
    {
        memset(b,0,sizeof(b));
        if(find(i,x))
            sum++;
    }
    return sum;
}
int main()
{
    int i,j,k,num;
    while(scanf("%d%d",&n,&m),n!=0||m!=0)
    {
        memset(map,0,sizeof(map));//图标用来记录气球分布
        memset(count,0,sizeof(count));//记数，记录各颜色的气球出现的次数
        memset(f,0,sizeof(f));//记录复合题意的气球
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            {
                scanf("%d",&map[i][j]);
                count[map[i][j]]++;
            }
            k=0;
            for(i=1;i<=50;i++)
            {
                if(count[i]&&met(i)>m)//如果i气球出现过，并且满足>m的条件
                    f[k++]=i;
            }
            if(k==0)
                printf("-1\n");
            else
            {
                for(i=0;i<k-1;i++)
                    printf("%d ",f[i]);
                printf("%d\n",f[k-1]);
            }
    }
    return 0;
}