Hdu1498



                                                 50 years, 50 colors

                                          Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                Total Submission(s): 2143    Accepted Submission(s): 1192

Problem Description

On Octorber 21st, HDU 50-year-celebration, 50-color balloons floating around the campus, it's so nice, isn't it? To celebrate this meaningful day, the ACM team of HDU hold some fuuny games. Especially, there will be a game named "crashing color balloons".

There will be a n*n matrix board on the ground, and each grid will have a color balloon in it.And the color of the ballon will be in the range of [1, 50].After the referee shouts "go!",you can begin to crash the balloons.Every time you can only choose one kind of balloon to crash, we define that the two balloons with the same color belong to the same kind.What's more, each time you can only choose a single row or column of balloon, and crash the balloons that with the color you had chosen. Of course, a lot of students are waiting to play this game, so we just give every student k times to crash the balloons.

Here comes the problem: which kind of balloon is impossible to be all crashed by a student in k times.

 

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
题意：
有一个n*n的矩阵，有k次操作，每次选一行或者一列，破坏其中的一种颜色的气球。问多少种颜色的气球还存在。
知道是二分图的问题，却无法想出解决的办法，翻了翻别人的博客，总算是看出来点眉目。
遍历每个颜色的气球，如果在图中出现过这种气球，找到这中气球的坐标（X,Y），将X,Y之间连一条线，
然后将问题变成求最小点覆盖，又因为最小点覆盖=最大匹配数，所以判断最大匹配数和k的关系。
代码：
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#define ll long long
#define MAXN 105
using namespace std;
int n,match[MAXN];
bool arc[MAXN][MAXN],inmap[MAXN],used[MAXN],s[MAXN][MAXN];
bool dfs(int x)
{
       int i;
       for (i=1;i<=n;i++)
          if (arc[x][i] && !used[i])
          {
                 used[i]=true;
                 if (!match[i] || dfs(match[i]))
                 {
                       match[i]=x;
                       return true;
                 }
          }
       return false;
}
int getmax()
{
       int sum=0;
       memset(match,0,sizeof(match));
       for (int i=1;i<=n;i++)
       {
               memset(used,false,sizeof(used));
               sum+=dfs(i);
       }
       return sum;
}
int main()
{
       int i,j,k,ans;
       bool f;
       while (~scanf("%d%d",&n,&k) && n)
       {
               memset(inmap,false,sizeof(inmap));
               for (i=1;i<=n;i++)
                  for (j=1;j<=n;j++)
                     scanf("%d",&s[i][j]),inmap[s[i][j]]=true;
               f=false;
               for (int t=1;t<=50;t++)
                  if (inmap[t])
                  {
                         memset(arc,false,sizeof(arc));
                         for (i=1;i<=n;i++)
                            for (j=1;j<=n;j++)
                                if (s[i][j]==t) arc[i][j]=true;
                         if (getmax()>k)
                         {
                                 if (!f) printf("%d",t);
                                    else printf(" %d",t);
                                 f=true;
                         }
                  }
               if (!f) printf("-1");
               printf("\n");
       }
       return 0;
}