50 years, 50 colors

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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：在校庆活动其中，有n*n个格子，里面放有不同颜色的汽球，让学生去拍汽球，但一次仅仅能拍一行或一列的同样颜色的汽球。而且一个学生最多仅仅能拍k次，问那些颜色的汽球不能在4k次中拍完则输出汽球的颜色，按升序排列，假设没有则输出－1.

解题：看起来无从下手，但是依据题意可知，我们能够把不同颜色的汽球分开来，对每种汽球各求一次。这样就是一个最小顶覆盖问题了。

#include<stdio.h>
#include<string.h>
int map[55][105][105],vist[105],match[105],n;
int find(int i,int c)
{
    for(int j=1;j<=n;j++)
    if(!vist[j]&&map[c][i][j])
    {
        vist[j]=1;
        if(match[j]==0||find(match[j],c))
        {
            match[j]=i; return 1;
        }
    }
    return 0;
}
int main()
{
    int m,c,cc[55];
    while(scanf("%d%d",&n,&m)>0&&n+m!=0)
    {
        memset(map,0,sizeof(map));
        memset(cc,0,sizeof(cc));
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            scanf("%d",&c); map[c][i][j]=1; cc[c]=1;
        }
        int flag=0,ans;
        for(int c=1;c<=50;c++)
        if(cc[c])
        {
            ans=0; memset(match,0,sizeof(match));
            for(int i=1;i<=n;i++)
            {
                memset(vist,0,sizeof(vist));
                ans+=find(i,c);
            }
            if(ans>m)
            {
               if(flag)printf(" "); printf("%d",c); flag=1;
            }
        }
        if(flag==0)printf("-1");
        printf("\n");
    }
}

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